
Q) Arithmetic mean is commonly known as_____________?
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Q) Arithmetic mean of a group of observations is defined as Mean = _________________________ / (Number of observations)?
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Q) A sequence of numbers is known as an______________?
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Q) Define Arithmetic Progression?
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Q) a_{n}+1  a_{n} = _______________?
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Q) The general form of an Arithmetic Progress is {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, .......then its nth term is a +___________?
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Q) nth term of an Arithmetic Progress from the end n^{th} term = First + (n  1) ___________ = a + (n  1) __________.?
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Q) Sum of First n Natural Numbers S=n/2(________+1) Using the formula S=n/2(a+l)?
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Q) Find the sum of first 35 natural numbers?
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Q) Find the sum of first 200 natural numbers?
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Q) Find the Difference and sum of first 500 natural numbers?
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Q) Find the sum of the squares of first 30 natural numbers.
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Q) Find the cubes of the squares of first 100 natural numbers?
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Q) Show that the sequence 8, 12, 16, 20, 24, ......... is an Arithmetic Progression. Find its 28^{th} term and the general term?
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Q) The 6^{th} term of an Arithmetic Progression is 17 and 14^{th} term of an Arithmetic Progression is 29. Find the first term and common difference of the Arithmetic Progression?
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Q) The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x?
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Q) A ladder has rungs 25 cm apart . The rungs decrease uniformly in length from 45 cm, at the bottom to 25 cm at the top. If the top and the bottom rungs are 2x(1/2) m apart, what is the length of the wood required for the rungs?
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Q) A small terrace at a football ground comprises of 1/5 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4m and a tread of 1/2m (see figure). Calculate the total volume of concrete required to build the terrace?
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 Arithmetic Progressions
 Arithmetic Mean
 General Form Of An Arithmetic Progress
 General Form Of An Arithmetic Progress
 Examples on Arithmetic Progression
 nth term of an Arithmetic Progress from the end
 examples on general form of an Arithmetic Progress
 Arithmetic Mean in Mathematics
 Examples on arithmetic means between the extreme terms
 Selection of Terms in an Arithmetic Progression
 examples to observe how to use the selection of terms in an arithmetic progression
 Properties of Arithmetic Progression
 Sum of the First n Terms of an Arithmetic Progression
 examples to find the sum of first n terms of an Arithmetic Progression
 Sum of the Cubes of First n Natural Numbers
 Examples to find the sum of the cubes of first n natural numbers
 Sum of First n Natural Numbers
 examples to find the sum of first n natural numbers
 Sum of the Squares of First n Natural Numbers
 examples to find the sum of the squares of first n natural numbers
 Arithmetic Progression Formulae
 Problems on Arithmetic Progression
 Problems on Sum of 'n' Terms of Arithmetic Progression
Definition of Arithmetic Mean (Average):
Arithmetic mean is commonly known as average. The average of a given set of numbers is called the arithmetic mean, or simply, the mean of the given numbers.
 Thus, the arithmetic mean of a group of observations is defined as
 Mean = (Sum of observations)/(Number of observations)
 x is the symbol of the arithmetic mean.
Thus, the mean of n observation x_{1}, x_{2}, . . .,x_{n}, is given by
Where the symbol ∑ called sigma, stands for summation, and we write,
 An arithmetic progression is a sequence of numbers in which the consecutive terms (beginning with the second term) are formed by adding a constant quantity with the preceding term.
 Definition of Arithmetic Progression: A sequence of numbers is known as an arithmetic progression (A.P.) if the difference of the term and the preceding term is always same or constant.
 The constant quantity stated in the above definition is called the common difference of the progression. The constant difference, generally denoted by d is called common difference.
 a_{n+1}  a_{n} = constant (=d) for all n∈ N
 From the definition, it is clear that an arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant.
 The general form of an Arithmetic Progress is {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ..........}, where ‘a’ is known as the first term of the Arithmetic Progress and ‘d’ is known as the common difference (C.D.).
 If a is the first term and d is the common difference of an Arithmetic Progress, then its nth term is
a + (n  1)d.  Let a_{1}, a_{2}, a_{3}, a_{4}, ........, a_{n}, .................. be the given Arithmetic Progress.
Then a_{1} = first term = a  By the definition, we have a_{2}  a_{1} = d
 Therefore, nth term of an Arithmetic Progress whose first term = ‘a’ and common difference = ‘d’ is
a_{n} = a + (n  1)d.
⇒ a_{2} = a_{1} + d
⇒ a_{2} = a + d
⇒ a_{2} = a +(2  1) d:
a_{3}  a_{2} = d
⇒ a_{3} = a_{2} + d
⇒ a_{3} = (a + d) + d
⇒ a_{3} = a + 2d
⇒ a_{3} = a + (3  1)d:
a_{4}  a_{3} = d
⇒ a_{4} = a_{3} + d
⇒ a_{4} = (a + 2d) + d
⇒ a_{4} = a + 3d
⇒ a_{4} = a +(4  1) d:
`a_5  a_4` = d
⇒a_{5} = `a_4` + d
⇒ a_{5} = (a + 3d) + d ⇒ a_{5} = a + 4d
⇒ a_{5} = a + (5  1)d:
Similarly, a_{6} = a + (6  1)d:
a_{7} = a +(7  1) d:
a_{n} = a + (n  1)d.
 2, 1, 4, 7, 10 ···· is an A.P. whose first term is 2 and find out common difference?
common difference is 1  (2) = 1 + 2 = 3.  The sequence {3, 7, 11, 15, 19, 23, 27,........} is an Arithmetic Progression. Find out the first term and common difference?
 whose common difference is 15,and firstterm=58 then find out the sequence of A.P?
 Find out the Arithmetic Progression whose common difference is 4,and first term=11?
First term=a=3,
difference=second termfirst term
d=73
d=4
Second term = First term (58) + (15)=43
Third term = Second term (43) + (15)=28
Fourth term = Third term (28) + (15)=13
Fifth term = Fourth term (13) + (15)=2 etc.
sequence {58, 43, 28, 13, 2, 17, 32, ········}
Second term (23) = First term (11) + 12=23
Third term (35) = Second term (23) + 12 = 35
Fourth term (47) = Third term (35) + 12 = 47
Fifth term (59) = Fourth term (47) + 12 = 59 etc.
sequence {11, 23, 35, 47, 59,.....}
Algorithm to determine whether a sequence is an Arithmetic Progression or not when its n^{th} term is given:
When a_{n+1} is independent of n then, the given sequence is a_{n} Arithmetic Progression. And, when a_{n+1} is not independent of n then, the given sequence is not an Arithmetic Progression.
1.Show that the sequence a_{n} defined by a_{n} = 2n + 3 is an Arithmetic Progression. Also find the common difference.
Solution:
The given sequence a_{n} = 2n + 3
Replacing n by (n + 1), we get
a_{n+1} = 2(n + 1) + 3
a_{n+1} = 2n + 2 + 3
a_{n+1} = 2n + 5
Now, a_{n+1}  a_{n} = (2n + 5)  (2n + 3) = 2n + 5  2n  3 = 2
Hence, a_{n+1}  a_{n} is independent of n, which is equal to 2.
Therefore, the given sequence a_{n} = 2n + 3 is an Arithmetic Progression with common difference 2.
2.Show that the sequence a_{n} defined by a_{n} = 3n^{2}+ 2 is not an Arithmetic Progression.
Solution:
The given sequence a_{n} = 3n^{2} + 2
Replacing n by (n + 1), we get
a_{n+1} = 3(n + 1)^{2} + 2
a_{n+1} = 3(n^{2} + 2n + 1) + 2
a_{n+1} = 3n^{2} + 6n + 3 + 2
a_{n+1} = 3n^{2} + 6n + 5
Now, a_{n+1}  a_{n} = (3n^{2} + 6n + 5)  (3n^{2} + 2) = 3n^{2} + 6n + 5  3n^{2}  2 = 6n + 3
Therefore, a_{n+1}  a_{n} is not independent of n.
Hence a_{n+1}  a_{n} is not constant.
Thus, the given sequence a_{n} = 3n^{2} + 2 is not an Arithmetic Progression.
Note:
To obtain the common difference of a given arithmetic progression we required to subtract its any term from that which follow it. That is,
 Let a and d be the first term and common difference of an Arithmetic Progress respectively having m terms.
 Then n^{th} term from the end is (m  n + 1)^{th} term from the beginning.
 Therefore, n^{th} term of the end = a_{m−n+1} = a + (m  n + 1  1)d = a + (m  n)d.
 We can also find the general term of an Arithmetic Progress according to the process below.
 To find the general term (or the n^{th} term) of the Arithmetic Progress {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ..........}.
 Clearly, for the Arithmetic Progress is {a, a + d, a + 2d, a + 3d, ..........} we have,
 Second term = a + d = a + (2  1)d = First term + (2  1) × Common Difference.
 Third term = a + 2d = a + (3  1)d = First term + (3  1) × Common Difference.
 Fourth term = a + 3d = a + (4  1)d = First term + (4  1) × Common Difference.
 Fifth term = a + 4d = a + (5  1)d = First term + (5  1) × Common Difference.
 Therefore, in general, we have,
 n^{th} term = First + (n  1) × Common Difference = a + (n  1) × d.
 Hence, if the n^{th} term of the Arithmetic Progress {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ..........} be denoted by tn, then t_{n} = a + (n  1) × d.
Show that the sequence 3, 5, 7, 9, 11, ......... is an Arithmetic Progress. Find its 15^{th} term and the general term.
Solution:
First term of the given sequence = 3
Second term of the given sequence = 5
Third term of the given sequence = 7
Fourth term of the given sequence = 9
Fifth term of the given sequence = 11
Now, Second term  First term = 5  3 = 2
Third term  Second term = 7  5 = 2
Fourth term  Third term = 9  7 = 2
Therefore, the given sequence is an Arithmetic Progress with the common difference 2.
We know that n^{th} term of an Arithmetic Progress, whose first term is a and common difference is d is t_{n} = a + (n  1) × d.
Therefore, 15^{th} term of the Arithmetic Progress = t_{15} = 3 + (15  1) × 2 = 3 + 14 × 2 = 3 + 28 = 31.
General term = n^{th} term = a_{n} = a + (n  1)d = 3 + (n  1) × 2 = 3 + 2n  2 = 2n + 1
Note:
General term = n^{th} term = a_{n} = a + (n  1)d
2.which term of the sequence 6, 11, 16, 21, 26, ....... is 126?
Solution:
First term of the given sequence = 6
Second term of the given sequence = 11
Third term of the given sequence = 16
Fourth term of the given sequence = 21
Fifth term of the given sequence = 26
Now, Second term  First term = 11  6 = 5
Third term  Second term = 16  11 = 5
Fourth term  Third term = 21  16 = 5
Therefore, the given sequence is an Arithmetic Progress with the common difference 5.
Let 126 is the n^{th} term of the given sequence. Then,
a_{n} = 126
⇒ a + (n  1)d = 126
⇒ 6 + (n  1) × 5 = 126
⇒ 6 + 5n  5 = 126
⇒ 5n + 1 = 126
⇒ 5n = 126  1
⇒ 5n = 125
⇒ n = 25
Hence, 25^{th} term of the given sequence is 126.
3.Find the 17^{th} term of the Arithmetic Progress {31, 25, 19, 13, ..................... }.
Solution:
The given Arithmetic Progress is {31, 25, 19, 13, ..................... }.
First term of the given sequence = 31
Second term of the given sequence = 25
Third term of the given sequence = 19
Fourth term of the given sequence = 13
Now, Second term  First term = 25  31 = 6
Third term  Second term = 19  25 = 6
Fourth term  Third term = 13  19 = 6
Therefore, common difference of the given sequence = 6.
Thus, the 17^{th} term of the given Arithmetic Progress
= a + (n 1)d
= 31 + (17  1) × (6)
= 31 + 16 × (6)
= 31  96
= 65.
Note:
 When given three quantities are in Arithmetic Progression, the middle one is known as the arithmetic mean of the other two.
 In the Arithmetic Progression {12, 22, 32}, 22 is the arithmetic mean between 12 and 32.
 In the Arithmetic Progression {7, 9, 11}, 9 is the arithmetic mean between 7 and 11.
 In the Arithmetic Progression {5, 6, 17}, 6 is the arithmetic mean between 5 and 17.
 In the Arithmetic Progression {8, 12, 16}, 12 is the arithmetic mean between 8 and 16.
 Let m be the arithmetic mean of two given quantities x and y. Then, x, m, y are in Arithmetic Progression.
 Now, m  x = y  m = common difference
⇒ 2m = x + y
⇒ `m = (x+y)/2`
 Therefore, the arithmetic mean between any two given quantities is half their sum.
 If more than three terms are in Arithmetic Progress, then the terms between the two extremes are called the arithmetic means between the extreme terms.
 In the Arithmetic Progression {3, 7, 11, 15, 19, 23, 27, 31, 35} the terms 7, 11, 15, 19, 23, 27 and 31 are the arithmetic means between the two extreme terms 3 and 35.
 In the Arithmetic Progression {5, 2, 1, 4, 7, 10, 13, 16, 19} the terms 2, 1, 4, 7, 10, 13 and 16 are the arithmetic means between the two extreme terms 5 and 19.
 In the Arithmetic Progression {85, 80, 75, 70, 65, 60, 55, 50, 45} the terms 80, 75, 70, 65, 60, 55, and 50 are the arithmetic means between the two extreme terms 85 and 45.
 If the sum of three terms in Arithmetic Progression be given, assume the numbers as a  d, a and a + d. Here common difference is d.
 If the sum of four terms in Arithmetic Progression be given, assume the numbers as a  3d, a  d, a + d and a + 3d.
 If the sum of five terms in Arithmetic Progression be given, assume the numbers as a  2d, a  d, a, a + d and a + 2d. Here common difference is 2d.
 If the sum of six terms in Arithmetic Progression be given, assume the numbers as a  5d, a  3d, a  d, a + d, a + 3d and a + 5d. Here common difference is 2d.
Note:
 From the above explanation we understand that in case of an odd number of terms, the middle term is ‘a’ and the common difference is ‘d’.
 Again, in case of an even number of terms the middle terms are a  d, a + d and the common difference is 2d.
 The sum of three numbers in Arithmetic Progression is 12 and the sum of their square is 56. Find the numbers.
Solution:
Let us assume that the three numbers in Arithmetic Progression be a  d, a and a + d.
According to the problem,
Sum = 12,
⇒ a  d + a + a + d = 12
⇒ 3a = 12
⇒ a = 4
and Sum of the squares = 56
(a  d)^{2} + a^{2} + (a + d)^{2} = 56
⇒ a^{2}  2ad + d^{2} + a^{2} + a^{2} + 2ad + d^{2} = 56
⇒ 3a^{2} + 2d^{2} = 56
⇒ 3 × (4)^{2} + 2d^{2} = 56 (∴a=4)
⇒ 3 × 16 + 2d^{2} = 56
⇒ 48 + 2d^{2} = 56
⇒ 2d^{2} = 56  48
⇒ 2d^{2} = 8
⇒ d^{2} = 4
⇒ d = ± 2
If d = 2, the numbers are 4 – 2, 4, 4 + 2 i.e., 2, 4, 6
If d = 2, the numbers are 4 + 2, 4, 4  2 i.e., 6, 4, 2
Therefore, the required numbers are 2, 4, 6 or 6, 4, 2.
2. The sum of four numbers in Arithmetic Progression is 20 and the sum of their square is 120. Find the numbers.
Solution:
Let us assume that the four numbers in Arithmetic Progression be a  3d, a  d, a + d and a + 3d.
According to the problem,
Sum = 20
⇒ a  3d + a  d + a + d + a + 3d = 20
⇒ 4a = 20
⇒ a = 5
and
and Sum of the squares = 120
⇒ (a  3d)^{2} + (a  d)^{2} + (a + d)^{2} + (a + 3d)^{2} = 120
⇒ a^{2}  6ad + 9d^{2} + a^{2}  2ad + d^{2} + a^{2} + 2ad + d^{2} + a^{2}+ 6ad + 9d^{2} = 120
⇒ 4a^{2} + 20d^{2} = 120
⇒ 4 × (5)^{2} + 20d^{2} = 120
⇒ 4 × 25 + 20d^{2} = 120
⇒ 100 + 20d^{2} = 120
⇒ 20d^{2} = 120  100
⇒ 20d^{2} = 20
⇒ d^{2} = 1
⇒ d = ± 1
If d = 1, the numbers are 5  3, 5  1, 5 + 1, 5 + 3 i.e., 2, 4, 6, 8
If d = 1, the numbers are 5 + 3, 5 + 1, 5  1, 5  3 i.e., 8, 6, 4, 2
Therefore, the required numbers are 2, 4, 6, 8 or 8, 6, 4, 2.
3. The sum of three numbers in Arithmetic Progression is 3 and their product is 8. Find the numbers.
Solution:
Let us assume that the three numbers in Arithmetic Progression be a  d, a and a + d.
According to the problem,
Sum = 3
⇒ a  d + a + a + d = 3
⇒3a = 3
⇒a = 1
and
Product = 8
⇒ (a  d) (a) (a + d) = 8
⇒ (1)[(1)^{2}  d^{2}] = 8
⇒ 1(1  d^{2}) = 8
⇒ 1 + d^{2} = 8
⇒ d^{2} = 8 + 1
⇒ d^{2} = 9
⇒ d = ± 3
If d = 3, the numbers are 1  3, 1, 1 + 3 i.e., 4, 1, 2
If d = 3, the numbers are 1 + 3, 1, 1  3 i.e., 2, 1, 4
Therefore, the required numbers are 4, 1, 2 or 2, 1, 4.
 We will discuss about some of the properties of Arithmetic Progression which we will frequently use in solving different types of problems on arithmetical progress.
Property I: If a constant quantity is added to or subtracted from each term of an Arithmetic Progression (A. P.), then the resulting terms of the sequence are also in A. P. with same common difference (C.D.).
Proof: Let {a_{1}, a_{2}, a_{3}, a_{4}, ..............} ........... (i) be an Arithmetic Progression with common difference d. Again, let k be a fixed constant quantity.
 Now k is added to each term of the above A.P. (i) Then the resulting sequence is a_{1} + k, a_{2} + k, a_{3} + k, a_{4} + k ..................
 Let b_{n} = a_{n} + k, n = 1, 2, 3, 4, ............ Then the new sequence is `b_1, b_2, b_3, b_4`, ...............
 We have b_{n+1}  b_{n} = (a_{n+1} + k)  (a_{n} + k) = a_{n+1}  a_{n} = d for all n ∈ N, [Since, n> is a sequence with common difference d].
 Therefore, the new sequence we get after adding a constant quantity k to each term of the A.P. is also an Arithmetic Progression with common difference d.
 Let us assume ‘a’ be the first term and ‘d’ be the common difference of an Arithmetic Progression. Then, the Arithmetic Progression is {a, a + d, a + 2d, a + 3d, a + 4d, ..........}
To get the clear concept of property I let us follow the below explanation.
1) By adding a constant quantity
 If a constant quantity k is added to each term of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, ............} we get, {a + k, a + d + k, a + 2d + k, a + 3d + k, a + 4d + k, ...........}............. (i)
 First term of the above sequence (i) is (a + k).
 Common difference of the above sequence (i) is (a + d + k)  (a + k) = d
 Therefore, the terms of the above sequence (i) form an Arithmetic Progression.
 Hence, if a constant quantity be added to each term of an Arithmetic Progression, the resulting terms are also in Arithmetic
 Progression with the same common difference.
2) By subtracting a constant quantity
 If a constant quantity k is subtracted from each term of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, ...........} we get, {a  k, a + d  k, a + 2d  k, a + 3d  k, a + 4d  k, ...........} ................. (ii)
 First term of the above sequence (ii) is (a  k). Common difference of the above sequence (ii) is (a + d  k)  (a  k) = d
 Therefore, the terms of the above sequence (ii) form an Arithmetic Progression.
 Hence, if a constant quantity be subtracted from each term of an Arithmetic Progression, the resulting terms are also in Arithmetic Progression with the same common difference.
Property II: If each term of an Arithmetic Progression is multiplied or divided by a nonzero constant quantity, then the resulting sequence form an Arithmetic Progression.
Proof:Let us assume {a_{1}, a_{2}, a_{3}, a_{4}, ..............} ........... (i) be an Arithmetic Progression with common difference d.
Again, let k be a fixed nonzero constant quantity.
Let us obtain, b_{1}, b_{2}, b_{3}, b_{4}, ............. be the sequence, after multiplying each term of the given A.P. (i) by k.
b_{1} = a_{1}k
b_{2} = a_{2}k
b_{3} = a_{3}k
b_{4} = a_{4}k
..............
..............
bn = a_{n}k
............
...........
To get the clear concept of property II let us follow the below explanation.
Let us assume ‘a’ be the first term and ‘d’ be the common difference of an Arithmetic Progression. Then, the Arithmetic Progression is {a, a + d, a + 2d, a + 3d, a + 4d, ............}
 If a nonzero constant quantity k (≠ 0) is multiplied by each term of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, ................} we get,
 {ak, ak + dk, ak + 2dk, ak + 3dk, .............} ............... (iii) First term of the above sequence (iii) is ak.
 Common difference of the above sequence (iii) is (ak + dk)  ak = dk
 Therefore, the terms of the above sequence (iii) form an Arithmetic Progression.
 Hence, if a nonzero constant quantity be multiplied by each term of an Arithmetic Progression, the resulting terms are also in Arithmetic Progression.
 If a nonzero constant quantity k (≠ 0) is divided by each term of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, ...........} we get,
 {a/k, a/k + d/k, a/k + 2d/k, a/k + 3d/k, ............} ................. (iv) First term of the above sequence (iv) is a/k.
 Therefore, the terms of the above sequence (iv) form an Arithmetic Progression.
 Hence, if a nonzero constant quantity be divided by each term of an Arithmetic Progression, the resulting terms are also in Arithmetic Progression.
Common difference of the above sequence (iv) is (a/k + d/k) – a/k = d/k
Property III:
In an Arithmetic Progression of finite number of terms the sum of any two terms equidistant from the beginning and the end is equal to the sum of the first and last terms.
 Let us assume ‘a’ be the first term, ‘d’ be the common difference, ‘l’ be the last term and ‘n’ be the number of terms of an A.P. (n is finite).
 The second term from the end = l  d
 The third term from the end = l  2d
The fourth term from the end = l  3d
The rth term from the end = l  (r  1)d
Again, the r^{th} term from the beginning = a + (r  1)d  Therefore, the sum of the r^{th} terms from the beginning a_{n} the end
= a + (r  1)d + l  (r  1)d
= a + rd  d + l  rd +d
= a + l  Hence, the sum of two terms equidistant from the beginning and the end is always same or equal to the sum of the first and last terms.
Property IV:
Three numbers x, y, and z are in Arithmetic Progression if and only if 2y = x + z.
Let us assume that x, y, z be in Arithmetic Progression.
Now, common difference = y  x and again, common difference = z  y
⇒ y  x = z  y
⇒2y = x + z
Conversely, let x, y, z be three numbers such that 2y = x + z. Then we prove that x, y, z are in Arithmetic Progression.
We have, 2y = x + z
⇒ y – x = z – y
⇒x, y, z are in Arithmetic Progression.
 A sequence is an Arithmetic Progression if and only if its n^{th} term is a linear expression in n i.e., a_{n} = An + B , where A, B are two constant quantities.
 In this case the coefficient of n in a_{n} is the common difference (C.D.) of the Arithmetic Progression.
Property VI:
 A sequence is an Arithmetic Progression if and only if the sum of its first n terms is of the form An^{2} + Bn, where A, B are two constant quantities that are independent of n.
 In this case the common difference is 2A that is 2 times the coefficient of n^{2}.
Property VII:
 A sequence is an Arithmetic Progression if the terms are selected at a regular interval from an Arithmetic Progression.
Property VIII:
 If x, y, and z are three consecutive terms of an Arithmetic Progression then 2y = x + z.
We will learn how to find the sum of first n terms of an Arithmetic Progression.
Prove that the sum S_{n} of n terms of an Arithmetic Progress (A.P.) whose first term ‘a’ and common difference ‘d’ is
S = n/2[2a + (n  1)d]
Or, S = n/2[a + l], where l = last term = a + (n  1)d.
Proof:
 Suppose, a_{n}, a_{n}, a_{n}, ………..be an Arithmetic Progression whose first term is a and common difference is d.
 Then, a_{1} = a
 a_{n} = a + (n  1)d
 S = a + (a + d) + (a + 2d) + (a + 3d) + ……….. + [a + (n  2)d] + [a + (n  1)d] ……………….. (i) By writing the terms of S in the reverse order, we get,
 S = [a + (n  1)d] + [a + (n  2)d] + [a + (n  3)d] + ……….+ (a + 3d) + (a + 2d) + (a + d) + a…… .(ii) Adding the corresponding terms of (i) and (ii), we get
 2S = [2a + (n  1)d] + [2a + (n  1)d] + [2a + (n  1)d] + ………. + [2a + (n  1)d]..n times
 2S = n[2a + (n 1)d]
⇒ `S = n/2[2a + (n  1)d]`
Now, l = last term = nth term = a + (n  1)d
Therefore, S = n/2[2a + (n  1)d]
=` n/2[a + {a + (n  1)d}]`
= `n/2[a + l]`.
a_{2} = a + d
a_{3} = a + 2d
a_{4} = a + 3d
………..
………..
Now,S = a_{1} + a_{2} + a_{3} + ………….. + a_{n1} + a_{n}
We can also find the sum of first n terms of an Arithmetic Progression according to the process below.
 Suppose, S denote the sum of the first n terms of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d ……………...}.
 Now n^{th} term of the given Arithmetic Progression is a + (n  1)d
 Let the n^{th} term of the given Arithmetic Progression = l Therefore, a + (n  1)d = l
 The term preceding the term (l  d) is l  2d and so on. Therefore, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. to n terms
 Or, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. + (l  2d) + (l  d) + l ……………… (i) Writing the above series in reverse order, we get
 S = l + (l  d) + (l  2d) + ……………. + (a + 2d) + (a + d) + a……………… (ii)
 Adding the corresponding terms of (i) and (ii), we get
 2S = (a + l) + (a + l) + (a + l) + ……………………. to n terms
⇒ 2S = n(a + l)
⇒ `S = n/2(a + l)`
⇒ `S = (Number of terms)/2 × (First term + Last term) `………… (iii)
⇒ `S = n/2[a + a + (n  1)d]`, Since last term l = a + (n  1)d
⇒ `S = n/2[2a + (n  1)d]`
Hence, the term preceding the last term is l – d.
1. Find the sum of the following Arithmetic series:
1 + 8 + 15 + 22 + 29 + 36 + ………………… to 17 terms
Solution:
First term of the given arithmetic series = 1
Second term of the given arithmetic series = 8
Third term of the given arithmetic series = 15
Fourth term of the given arithmetic series = 22
Fifth term of the given arithmetic series = 29
Now, Second term  First term = 8  1 = 7
Third term  Second term = 15  8 = 7
Fourth term  Third term = 22  15 = 7
 Therefore, common difference of the given arithmetic series is 7.
 The number of terms of the given A. P. series (n) = 17
 We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is S = n/2[2a + (n  1)d]
 Therefore, the required sum of first 20 terms of the series =` (17)/2[2 ** 1 + (17  1) ** 7]`
= `17/2[2 + 112]`
= `17/2 × 114`
= 17 × 57
= 969
2. Find the sum of the series: 7 + 15 + 23 + 31 + 39 + 47 + ……….. + 255
Solution:
First term of the given arithmetic series = 7
Second term of the given arithmetic series = 15
Third term of the given arithmetic series = 23
Fourth term of the given arithmetic series = 31
Fifth term of the given arithmetic series = 39
Now, Second term  First term = 15  7 = 8
Third term  Second term = 23  15 = 8
Fourth term  Third term = 31  23 = 8
Therefore, the given sequence is an arithmetic series with the common difference 8.
Let there be n terms in the given arithmetic series. Then
a_{n} = 255
⇒ a + (n  1)d = 255
⇒ 7 + (n  1) × 8 = 255
⇒ 7 + 8n  8 = 255
⇒ 8n  1 = 255
⇒ 8n = 256
⇒ n = 256/8
⇒ n = 32
Therefore, the required sum of the series = `32/2[2 ** 7 + (32  1) **8]`
= 16 [14 + 31 ∙ 8]
= 16 [14 + 248]
= 16 × 262
= 4192
 We know the formula to find the sum of first n terms of an Arithmetic Progression is S =`n/2[2a + (n  1)d]`.
In the formula there are four quantities. They are S, a, n and d. If any three quantities are known, the fourth quantity can be determined.
Suppose when two quantities are given then, the remaining two quantities are provided by some other relation.  When the sum S_{n} of n terms of an Arithmetic Progression is given, then the nth term `a_n` of the Arithmetic Progression cann be determined by the formula a_{n} = S_{n}  S_{n−1}.
Let us assume the required sum = S.
Therefore, S = 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ................... + n^{3}
Now, we will use the below identity to find the value of S:
Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get
`2^4  1^4 =(4 xx 2^3)(6 xx 2^2)+(4 xx 2)1`
`3^4  2^4 =(4 xx 3^3)(6 xx 3^2)+(4 xx 3)1`
`4^4  3^4 =(4 xx 4^3)(6 xx 4^2)+(4 xx 4)1`
_______________________________________________
`n^4 (n  1)^4 =(4 xx n^3)(6 xx n^2)+(4 xx n)1`
Adding we get,
n^{4}  `0^4` = 4(1^{3} + 2^{3} + 3^{3} + 4^{3} + ...... + n^{3})  6(1^{2} + 2^{2} + 3^{2} + 4^{2}+ ........ + n^{2}) + 4(1 + 2 + 3 + 4 + ........ + n)  (1 + 1 + 1 + 1 + ......... n times)
⇒ n^{4} = 4S  6 . n(n + 1)(2n + 1)/6 + 4 ∙ n(n+1)/2  n
⇒ n^{4} + n(n + 1)(2n + 1)  2n(n + 1) + n = 4S
⇒ 4S = n^{4} + n(n + 1)(2n + 1)  2n(n + 1) + n
⇒ 4S = n^{4} + n(2n^{2} + 3n + 1) – 2n^{2}  2n + n
⇒ 4S = n^{4} + 2n^{3} + 3n^{2} + n  2n^{2}  2n + n
⇒ 4S = n^{4} + 2n^{3} + n^{2}
⇒ 4S = n^{2}(n^{2} + 2n + 1)
⇒ 4S = n^{2}(n + 1)^{2}
Therefore, `S = (n^2(n+1)^2)/4 = [(n(n+1))/2]^2` = (Sum of the first n natural numbers)^{2}
i.e., 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ................... + n^{3} = `[(n(n+1))/2]^2`
Thus, the sum of the cubes of first n natural numbers = `[(n(n+1))/2]^2`
1. Find the sum of the cubes of first 12 natural numbers.
Solution:
Sum of the cubes of first 12 natural numbers
i.e., 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ................... + 12^{3}
We know the sum of the cubes of first n natural numbers (S) = `[(n(n+1))/2]^2`
Here n = 12
Therefore, the sum of the cubes of first 12 natural numbers = `[(12(12+1))/2]^2`
= `[(12×13)/2]^2`
= `[6 × 13]^2`
= `(78)^2`
= `6084`
2. Find the sum of the cubes of first 25 natural numbers.
Solution:
Sum of the cubes of first 25 natural numbers
i.e., 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ................... + 25^{3}
We know the sum of the cubes of first n natural numbers (S) = `[(n(n+1))/2]^2`
Here n = 25
Therefore, the sum of the cubes of first 25 natural numbers = `[(25(25+1))/2]^2`
= `[(25×26)/2]^2`
= `[25 × 13]^2`
= `(325)^2`
= `105625`
 We will discuss here how to find the sum of first n natural numbers.
 Let S be the required sum. Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + n
 Clearly, it is an Arithmetic Progression whose first term = 1, last term = n and number of terms = n.
 Therefore, `S = n/2(n + 1)`, [Using the formula `S = n/2(a + l)`]
 Find the sum of first 25 natural numbers.
Solution: Let S be the required sum.
Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 25
Clearly, it is an Arithmetic Progression whose first term = 1, last term = 25 and number of terms = 25.
Therefore, `S = (25)/2(25 + 1)`, [Using the formula `S = n/2(a + l)`]
= `25/2(26)`
= 25 × 13
= 325
Therefore, the sum of first 25 natural numbers is 325.
 Find the sum of first 100 natural numbers.
Solution:
Let S be the required sum.
Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 100
Clearly, it is an Arithmetic Progression whose first term = 1, last term = 100 and number of terms = 100.
Therefore, `S = (100)/2 (100 + 1)`, [Using the formula `S = n/2(a + l)`]
= 50(101)
= 5050
Therefore, the sum of first 100 natural numbers is 5050.
 Find the sum of first 500 natural numbers.
Solution:
Let S be the required sum.
Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 500
Clearly, it is an Arithmetic Progression whose first term = 1, last term = 500 and number of terms = 500.
Therefore, `S = (500)/2(500 + 1)`, [Using the formula `S = n/2(a + l)`]
= 225(501)
= 112725
Therefore, the sum of first 500 natural numbers is 112725.
Let us assume the required sum = S
Therefore, S = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + ................... + n^{2}
Now, we will use the below identity to find the value of S:
Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get
` 2^3  1^3 = 3 xx 2^2  3 xx 2 + 1`
`3^3  2^3 = 3 xx 3^2  3 xx 3 + 1`
`4^3  3^3 = 3 xx 4^2  3 xx 4 + 1`
__________________________________
`n^3  (n  1)^3 = 3 xx n^2  3 xx n + 1`
Adding we get, `n^3  0^3 = 3(1^2 + 2^2 + 3^2 + 4^2 + ........... + n^2) `
`3(1 + 2 + 3 + 4 + ........ + n) + (1 + 1 + 1 + 1 + ......... n times)`
⇒ n^{3} = 3S  `3 ( [n(n+1)]/2 )`+ n
⇒3S = n^{3} +` [3n(n+1)]/2` – n
3S= n(n^{2}  1) + ` [3n(n+1)]/2`
3S= n(n^{2}  1) + ` [3n(n+1)]/2`
⇒ 3S = n(n + 1)(n  1 + 3/2) (since, n^{2}  1=(n1)(n+1) and n(n+1) is common in both terms)
⇒ 3S = n(n + 1)`[(2n−2+3)/2]`
⇒ `3S = [n(n+1)(2n+1)]/2`
Therefore,` S = [n(n+1)(2n+1)]/6`
find the sum of the squares of first 50 natural numbers.
Solution:
We know the sum of the squares of first n natural numbers (S) = `[n(n+1)(2n+1)]/6`
Here n = 50
Therefore, the sum of the squares of first 50 natural numbers = `[50(50+1)(2×50+1)]/6`
= `[50×51×101]/6`
=` [257550]/6`
= 42925
 Find the sum of the squares of first 100 natural numbers.
Solution: We know the sum of the squares of first n natural numbers (S) = `[n(n+1)(2n+1)]/6`
Here n = 100
Therefore, the sum of the squares of first 50 natural numbers = `[100(100+1)(2×100+1)]/6`
= `[100×101×201]/6`
= `[2030100]/6`
= 338350
 Let ‘a’ be the first term and ‘d’ the common difference of an Arithmetic Progression. Then its
General term (`n^(th)` term) = tn = a + (n  1)d
(ii) The sum of the first n terms = Sn = n(a + l)/2 = Number of terms(First term + Last term)/2  (i) The sum of first n natural numbers (Sn)= `[n(n + 1)]/2`
 The arithmetic mean between two given quantities a and b = `1/2`× (Sum of the given quantities) = `(a+b)/2`.
 (i) If the sum of three terms in Arithmetic Progression be given, assume the numbers as a  d, a and a + d. Here common difference is d.
or, Sn = n/2[2a + (n  1)d] where l = last term = nth term = a + (n  1)d.
i.e., 1 + 2 + 3 + 4 + 5 + .................... + n = `[n(n + 1)]/2`
(ii) The sum of the squares of first n natural numbers (Sn)= `[n(n + 1)(2n + 1)]/6`
i.e.,1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + ................... + n^{2} =` [n(n + 1)(2n +1)]/6`
(iii) The sum of the cubes of first n natural numbers (Sn) =` [(n(n+1))/2]^2`
i.e., 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ........... + n^{3}=`[(n(n+1))/2]^2`
(ii) If the sum of four terms in Arithmetic Progression be given, assume the numbers as a  3d, a  d, a + d and a + 3d.
(iii) If the sum of five terms in Arithmetic Progression be given, assume the numbers as a  2d, a  d, a, a + d and a + 2d. Here common difference is 2d.
(iv) If the sum of six terms in Arithmetic Progression be given, assume the numbers as a  5d, a  3d, a  d, a + d, a + 3d and a + 5d. Here common difference is 2d.
Here we will learn how to solve different types of problems on arithmetic progression.
 Show that the sequence 7, 11, 15, 19, 23, ......... is an Arithmetic Progression. Find its 27^{th} term and the general term.
Solution:
First term of the given sequence = 7
Second term of the given sequence = 11
Third term of the given sequence = 15
Fourth term of the given sequence = 19
Fifth term of the given sequence = 23
Now, Second term  First term = 11  7 = 4
Third term  Second term = 15  11 = 4
Fourth term  Third term = 19  15 = 4
Fifth term – Fourth term = 23  19 = 4
Therefore, the given sequence is an Arithmetic Progress with the common difference 4.
We know that n^{th} term of an Arithmetic Progress, whose first term is a and common difference is d is t_{n} = a + (n  1) × d.
Therefore, 27^{th} term of the Arithmetic Progress = t_{n} = 7 + (27  1) × 4 = 7 + 26 × 4 = 7 + 104 = 111.
General term = nth term = a_{n} = a + (n  1)d = 7 + (n  1) × 4 = 7 + 4n  4 = 4n + 3
 The 5^{th} term of an Arithmetic Progression is 16 and 13^{th} term of an Arithmetic Progression is 28. Find the first term and common difference of the Arithmetic Progression. Solution:
Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the required Arithmetic Progression. According to the problem,
5^{th} term of an Arithmetic Progression is 16
i.e., 5^{th} term = 16
⇒ a + (5  1)d = 16
⇒ a + 4d = 16 ................... (i)
and 13^{th} term of an Arithmetic Progression is 28
i.e., 13^{th} term = 28
⇒ a + (13  1)d = 28
⇒ a + 12d = 28 .................... (ii)
Now, subtract the equation (i) from (ii) we get,
8d = 12
⇒ d = 12/8
⇒ d = 3/2
Substitute the value of d = 3/2 in the equation (i) we get,
⇒ a + 4 × 3/2 = 16
⇒ a + 6 = 16
⇒ a = 16  6
⇒ a = 10
Therefore, the first term of the Arithmetic Progression is 10 and common difference of the Arithmetic Progression is 3/2.
 Find the sum of the first 35 terms of an Arithmetic Progression whose third term is 7 and seventh term is two more than thrice of its third term.
Solution:
Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the given Arithmetic Progression.
According to the problem,
3rd term of an Arithmetic Progression is 7
i.e., 3^{th} term = 7
⇒ a + (3  1)d = 7
⇒ a + 2d = 7 ................... (i)
and seventh term is two more than thrice of its third term.
i.e., 7th term = 3 × 3^{rd} term + 2
⇒ a + (7  1)d = 3 × [a + (3  1)d] + 2
⇒ a + 6d = 3 × [a + 2d] + 2
Substitute the value of a + 2d = 7 we get,
⇒ a + 6d = 3 × 7 + 2
⇒ a + 6d = 21 + 2
⇒ a + 6d = 23 ................... (ii)
Now, subtract the equation (i) from (ii) we get,
4d = 16
⇒ d = 16/4
⇒ d = 4
Substitute the value of d = 4 in the equation (i) we get,
⇒ a + 2 × 4 = 7
⇒ a + 8 = 7
⇒ a = 7  8
⇒ a = 1
Therefore, the first term of the Arithmetic Progression is 1 and common difference of the Arithmetic Progression is 4.
Now, sum of the first 35 terms of an Arithmetic Progression `S^35=(35)/2[2 × (1) + (35  1) × 4]`, [Using the Sum of the First n Terms of an Arithmetic Progression S_{n} = `n/2[2a + (n  1)d]` = `(35)/2[2 + 34 × 4]`
= `(35)/2[2 + 136]`
= `(35)/2[134]`
= `35 × 67`
`S_35= 2345`.
 If the 5^{th} term and 12^{th} term of an Arithmetic Progression are 30 and 65 respectively, find the sum of its 26 terms.
Solution:
Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the given Arithmetic Progression.
According to the problem,
5^{th} term of an Arithmetic Progression is 30
i.e., 5^{th} term = 30
⇒ a + (5  1)d = 30
⇒ a + 4d = 30 ................... (i)
and 12^{th} term of an Arithmetic Progression is 65
i.e., 12^{th} term = 65
⇒ a + (12  1)d = 65
⇒ a + 11d = 65 .................... (ii)
Now, subtract the equation (i) from (ii) we get,
7d = 35
⇒` d = (35)/7`
⇒` d = 5`
⇒Substitute the value of d = 5 in the equation (i) we get,
a + 4 × 5 = 30
⇒ a + 20 = 30
⇒ a = 30  20
⇒ a = 10
 Therefore, the first term of the Arithmetic Progression is 10 and common difference of the Arithmetic Progression is 5.
 Now, sum of the first 26 terms of an Arithmetic Progression `S_26 = (26)/2[2 × 10 + (26  1) × 5]`, [Using the Sum of the First n Terms of an Arithmetic Progression` S_n = n/2[2a + (n  1)d]`
= `13[20 + 25 × 5]`
= `13[20 + 125]`
= `13[145]`
= `1885`
 Find the arithmetic mean of the first ten natural numbers
 Find the arithmetic mean of the first eight odd numbers
 Find the arithmetic mean of the first seven multiples of 5
 Find the arithmetic mean of all the factors of 20
 Find the arithmetic mean of the first ten prime numbers
 Find the arithmetic mean of the first eleven even numbers
 The number of members in 10 families of a society are 2, 4, 3, 4, 2, 2, 3, 5, 1, 4. Find the mean number of members per family.
 The following are the number of maths book issued in a school library during a week: 222, 120, 201, 322, 167, 273, 406 and 345 Find the average number of maths book issued per day.
 If the arithmetic mean of 7, 9, 11, 13, x, 21 is 13, find the value of x.
 The daily minimum temperature recorded (in degree F) at a place during a week was as under:
Saturday Monday Tuesday Wednesday Thursday Friday 34.5 35.8 38.3 37.8 35.1 33.9  The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?
 The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean?
 The mean of 15 numbers is 27. If each number is multiplied by 5, what will be the mean of the new numbers?
 The mean of 15 numbers is 63. If each number is divided by 9, what will be the mean of the new numbers?
 The percentages of marks obtained by 12 students of a class in mathematics are 36, 64, 47, 43, 50, 39, 81, 93, 72, 35, 53, 41. Find the mean percentage of marks.
 The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of the new set of 20 numbers.
 The mean weight of 8 girls in a group is 35 kg. The individual weights of seven of them are 35 kg, 28 kg, 43 kg, 25 kg, 33 kg, 51 kg and 22 kg. Find the weight of the eighth girls.
 The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 75 was misread as 25. Find the correct mean.
 The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. Find the correct mean.
 The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.
Answers for the worksheet on arithmetic mean are given below to check the exact of the above questions on mean (average).
 5.5
 8
 20
 7
 12.9
 12
 3
 257
 x = 17
 35.4 °F
 38
 37
 135
 7
 54.5
 19.5
 43
 40
 64.91
 38
 Consider the following lists of numbers :
(i) 1, 2, 3, 4, . . .
(ii) 100, 70, 40, 10, . . .
(iii) – 3, –2, –1, 0, . . .
(iv) 3, 3, 3, 3, . . .
(v) –1.0, –1.5, –2.0, –2.5, . . .
 Each of the numbers in the list is called a term.
 In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression ( AP ).
 So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
 This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero.
Example 1:
For the AP : 3/2, 1/2, –1/2, –3/2 , . . ., write the first term a and the common difference d.
Solution :
Here, a = 3/2 ,
d = 1/2 – 3/2 = – 1 the numbers are in AP.
Example 2 : Which of the following list of numbers form an AP?
If they form an AP, write the next two terms:
(i) 4, 10, 16, 22, ... .. .. ......(ii) 1, – 1, – 3, – 5, . . .
(iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .
Solution :
(i) We have a_{2}– a_{1} = 10 – 4 = 6
a_{3} – a_{2} = 16 – 10 = 6
a_{4} – a_{3} = 22 – 16 = 6
i.e., a_{k + 1} – a_{k} is the same every time.
The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.
(ii) a_{2}– a_{1} = – 1 – 1 = – 2
a_{3} – a_{2} = – 3 – ( –1 ) = – 3 + 1 = – 2
a_{4}– a_{3} = – 5 – ( –3 ) = – 5 + 3 = – 2
i.e., a_{k + 1} – a_{k} is the same every time.
The next two terms are: – 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9
(iii) a_{2}– a_{1} = 2 – (– 2) = 2 + 2 = 4
a_{3} – a_{2} = – 2 – 2 = – 4
As a_{2}– a_{1} ≠ a_{3} – a_{2} , the given list of numbers does not form an AP.
(iv) a_{2} – a_{1} = 1 – 1 = 0
a_{3} – a_{2} = 1 – 1 = 0
a_{4} – a_{3} = 2 – 1 = 1
Here, a_{2} – a_{1} = a_{3}– a_{2} ≠ a_{4} – a_{3}.
So, the given list of numbers does not form an AP.
 Let us consider the situation, in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of 8000, with an annual increment of 500. What would be her monthly salary for the fifth year?
 To answer this, let us first see what her monthly salary for the second year would be.
 It would be (8000 + 500) = 8500. In the same way, we can find the monthly salary for the 3^{rd}, 4^{th} and 5^{th} year by adding 500 to the salary of the previous year. So, the salary for the 3^{rd} year = (8500 + 500)
 Observe that we are getting a list of numbers 8000, 8500, 9000, 9500, 10000, .. These numbers are in AP
 First salary + (15 – 1) × Annual increment.
 In the same way, her monthly salary for the 25^{th} year would be
 Rs [8000 + (25 – 1) × 500] = 20000 = First salary + (25 – 1) × Annual increment
 the second term a_{2} = a + d = a + (2 – 1) d
 the third term a_{3}= a_{2}+ d = (a + d) + d = a + 2d = a + (3 – 1)d
 the fourth term a_{4} = a_{3}+ d = (a + 2d) + d = a + 3d = a + (4 – 1)d.
 the nth term a_{n} = a + (n – 1)d.
= (8000 + 500 + 500)
= (8000 + 2 × 500)
= [8000 + (3 – 1) × 500] (for the 3^{rd} year)
= Rs 9000
Salary for the 4^{th} year = (9000 + 500)
= Rs (8000 + 500 + 500 + 500)
= (8000 + 3 × 500)
= [8000 + (4 – 1) × 500] (for the 4^{th} year)
= Rs 9500
Salary for the 5^{th} year = (9500 + 500)
= (8000 + 500 + 500 + 500 + 500)
= (8000 + 4 × 500)
= [8000 + (5 – 1) × 500] (for the 5^{th} year)
= 10000
Salary for the 15^{th} year= Salary for the 14^{th} year + 500
= [8000 + 14 × 500]
= [8000 + (15 – 1) × 500] = Rs 15000
Let a_{1}, a_{2}, a_{3}, . . . be an AP whose first term a_{1} is a and the common difference is d.
a_{3} = a + (3 – 1)d
a_{4} = a + (4 – 1)d
a_{n} = a + (n – 1)d.
a_{n}= a + (n – 1)d
Example 3 : Find the 10^{th} term of the AP : 2, 7, 12, . . .. .. .
Solution:
Here, a = 2, d = 7 – 2 = 5 and n = 10.
We have a_{n} = a + (n – 1) d
So, a_{10} = 2 + (10 – 1) × 5 = 2 + 45 = 47
Therefore, the 10^{th} term of the given AP is 47.
Example 4 : Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0?
Solution :
Here, a = 21, d = 18 – 21 = – 3 and a_{n} = – 81, and we have to find n.
As a_{n} = a + ( n – 1) d,
we have – 81 = 21 + (n – 1)(– 3)
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35
 Therefore, the 35^{th} term of the given AP is – 81.
 Next, we want to know if there is any n for which a_{n} = 0. If such an n is there, then 21 + (n – 1) (–3) = 0,
i.e., n =8
So, the eighth term is 0
Example 5 :Determine the AP whose 3^{rd} term is 5 and the 7^{th} term is 9.
Solution :
We have a_{3} = a + (3 – 1) d = a + 2d = 5 ...........(1)
and a_{7}= a + (7 – 1) d = a + 6d = 9 ............(2)
Solving the pair of linear equations (1) and (2), we get
a = 3, d = 1
Hence, the required AP is 3, 4, 5, 6, 7, . . .
Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .
Solution :
We have a_{2}– a_{1} = 11 – 5 = 6,
a_{3}– a_{2}= 17 – 11 = 6
a_{4}– a_{3} = 23 – 17 = 6
As a_{k + 1} – a_{k} is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
Now, a = 5 and d = 6.
Let 301 be a term, say, the n^{th} term of the this AP.
a_{n}= a + (n – 1) d
So, 301 = 5 + (n – 1) × 6
i.e., 301 = 6n – 1
So, n =302/6= 151/3
 But n should be a positive integer. So, 301 is not a term of the given list of numbers.
Example 7 : How many twodigit numbers are divisible by 3?
Solution :
The list of twodigit numbers divisible by 3 is 12, 15, 18, . . . , 99 Is this an AP
Here, a = 12, d = 3, a_{n}= 99.
As a_{n} = a + (n – 1) d,
we have 99 = 12 + (n – 1) × 3
i.e., 87 = (n – 1) × 3
n – 1 =87/3= 29
i.e., n = 29 + 1 = 30
So, there are 30 twodigit numbers divisible by 3.
Example 8: Find the 11^{th} term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62.
Solution :
Here, a = 10, d = 7 – 10 = – 3, l = – 62,
where l = a + (n – 1)d
 11^{th} term from the last term, we will find the total number of terms in the AP.
i.e., – 72 = (n – 1)(–3)
i.e., n – 1 = 24
or n = 25
So, a_{15} = 10 + (15 – 1)(–3) = 10 – 42 = – 32
i.e., the 11^{th} term from the last term is – 32.
 If we write the given AP in the reverse order, then a = – 62 and d = 3
So, the question now becomes finding the 11^{th} term with these a and d.
So, a_{11} = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32
So, the 11^{th} term, which is now the required term, is – 32.
Example 9: A sum of 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
Solution:
We know that the formula to calculate simple interest is given by Simple Interest =`(PxxRxxT)/100`
 So, the interest at the end of the 1^{st} year =`(1000 xx 8 xx 1)/100`= Rs 80
 The interest at the end of the 2^{nd} year =`(1000 xx 8 xx 2)/100` =Rs 160
 The interest at the end of the 3^{rd} year =`(1000 xx 8 xx 3)/100` = Rs 240
 Similarly, we can obtain the interest at the end of the 4^{th} year, 5^{th} year, and so on.
 So, the interest (in Rs) at the end of the 1^{st}, 2^{nd}, 3^{rd}, . . . years, respectively are 80, 160, 240, . . .
 It is an AP as the difference between the consecutive terms in the list is 80, i.e., d = 80. Also, a = 80.
 So, to find the interest at the end of 30 years.
 Now, a_{30} = a + (30 – 1) d = 80 + 29 × 80 = 2400
 So, the interest at the end of 30 years will be Rs 2400.
Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Solution:
The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :
23, 21, 19, . . ., 5
It forms an AP .
 Let the number of rows in the flower bed be n.
 Then a = 23, d = 21 – 23 = – 2, a_{n} = 5 As, a_{n}= a + (n – 1) d
 So, there are 10 rows in the flower bed.
 Here, a = 100, d = 50 and n = 21.
 Using the formula :S = n/2(2a+(n1)d) , S=21/2 (2x100 +21 1) x50)
We have, 5 = 23 + (n – 1)(– 2)
i.e., – 18 = (n – 1)(– 2)
i.e., n = 10
=21/2(200+1000)
=21/2x1200=12600
So, the amount of money collected on her 21^{st} birthday is Rs 12600.
Remark :
The n^{th} term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., an= S_{n} – S_{n – 1}
Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .
Solution:
Here, a = 8, d = 3 – 8 = –5, n = 22.
We know that
`S =n/2[2a+(n1)d]`
Therefore,` S = (22)/2[16+21(5)]`
=` 11(–89) = – 979`
So, the sum of the first 22 terms of the AP is – 979
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution: Here, S_{14} = 1050, n = 14, a = 10.
As `S_n=n/2[2a+(n1)d]`
`1050=(14)/2[20+13d]=140+91d`
`910 = 91d`
`d = 10`
Therefore, `a_20 = 10 + (20 – 1) × 10 = 200`
`20^(th) term is 200`.
Example 13 : How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?
Solution: Here, a = 24, d = 21 – 24 = –3,
S_{n}= 78. We need to find n.
We know that `S_n=n/2[2a+(n1)d]`
`78 = n/2[48 (n1)( 3)]`
`78=n/ 2[513n]`
Or `3n^2– 51n + 156 = 0`
or `n^2 – 17n + 52 = 0`
or `(n – 4)(n – 13) = 0`
or `n = 4 or 13`
Example 14 : Find the sum of:
(i) the first 1000 positive integers (ii) the first n positive integers
Solution :
(i) Let S = 1 + 2 + 3 + . . . + 1000
Using the formula S_{n} =n/2(a+1) for the sum of the first n terms of an AP, we have
S_{1000} =1000/2 (1+ 1000)
= 500 × 1001 = 500500
So, the sum of the first 1000 positive integers is 500500.
(ii) Let S_{n} = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
`S_n=(n(1+n))/2 or S_n = (n+1)/2`
So, the sum of first n positive integers is given by
`S_n =(n(n+1))/2`
Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given bya_{n} = 3 + 2n.
Solution :
a_{n}= 3 + 2n,
so, a_{1}= 3 + 2 = 5
a_{2} = 3 + 2 × 2 = 7
a_{3} = 3 + 2 × 3 = 9
List of numbers becomes 5, 7, 9, 11, . . .
Here, 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
So, it forms an AP with common difference d = 2.
To find S_{24} , we have n = 24, a = 5, d = 2.
`S_24 = (24)/ 2[2xx 5 +(24 1)xx 2]`
= `12[ 10+46] = 672`.
 So, sum of first 24 terms of the list of numbers is 672.
Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find
(i) the production in the 1st year (ii) the production in the 10th year
(iii) the total production in first 7 years
Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1^{st}, 2^{nd}, 3^{rd}, . . ., years will form an AP.
Let us denote the number of TV sets manufactured in the n^{th} year by an.
Then, a_{3} = 600 and
a_{7}= 700
a + 2d = 600
and a + 6d = 700
So, production of TV sets in the 10^{th} year is 775.
(iii) Also, `S_7 = 7/ 2[2xx 550+ (7 1)xx 25]` =`7/2[ 1100 +150]`
=4375
1. Arithmetic mean is commonly known as average.
2. The average of a given set of numbers is called the arithmetic mean, or simply, the mean of the given numbers.
3. Thus, the arithmetic mean of a group of observations is defined as
Mean = (Sum of observations)/(Number of observations)
x is the symbol of the arithmetic mean.
`bar(x)`= { x_{1} +x_{2} +x_{3} +.........+x_{n} } / n
Definition of Arithmetic Progression:
A sequence of numbers is known as an arithmetic progression (A.P.) if the difference of the term and the preceding term is always same or constant.
a_{n+1}  a_{n} = constant (=d) for all n∈ N
f a is the first term and d is the common difference of an Arithmetic Progress, then its nth term is
a + (n  1)d.
Let a and d be the first term and common difference of an Arithmetic Progress respectively having m terms.
 Then n^{th} term from the end is (m  n + 1)^{th} term from the beginning.
 Therefore, n^{th} term of the end = a_{m−n+1} = a + (m  n + 1  1)d = a + (m  n)d.
Sum of First n Natural Numbers
We will discuss here how to find the sum of first n natural numbers.
Let S be the required sum.
Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + n
Clearly, it is an Arithmetic Progression whose first term = 1, last term = n and number of terms = n.
Therefore, `S = n/2(n + 1)`, [Using the formula `S = n/2(a + l)`]
Sum of the Squares of First n Natural Numbers
the sum of the squares of first n natural numbers
S = `(n(n+1)(2n+1))/6`
 Let ‘a’ be the first term and ‘d’ the common difference of an Arithmetic Progression. Then its
General term (nth term) = tn = a + (n  1)d
(ii) The sum of the first n terms = Sn = n(a + l)/2 = Number of terms(First term + Last term)/2  (i) The sum of first n natural numbers (Sn)= `[n(n + 1)]/2`
 The arithmetic mean between two given quantities a and b = `1/2`× (Sum of the given quantities) = `(a+b)/2`.
 (i) If the sum of three terms in Arithmetic Progression be given, assume the numbers as a  d, a and a + d. Here common difference is d.
or, Sn = n/2[2a + (n  1)d] where l = last term = nth term = a + (n  1)d.
i.e., 1 + 2 + 3 + 4 + 5 + .................... + n = `[n(n + 1)]/2`
(ii) The sum of the squares of first n natural numbers (Sn)= `[n(n + 1)(2n + 1)]/6`
i.e.,1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + ................... + n^{2} =` [n(n + 1)(2n +1)]/6`
(iii) The sum of the cubes of first n natural numbers (Sn) =` [(n(n+1))/2]^2`
i.e., 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ........... + n^{3}=` [(n(n+1))/2]^2`
(ii) If the sum of four terms in Arithmetic Progression be given, assume the numbers as a  3d, a  d, a + d and a + 3d.
(iii) If the sum of five terms in Arithmetic Progression be given, assume the numbers as a  2d, a  d, a, a + d and a + 2d. Here common difference is 2d.
(iv) If the sum of six terms in Arithmetic Progression be given, assume the numbers as a  5d, a  3d, a  d, a + d, a + 3d and a + 5d. Here common difference is 2d.