Sum of Arithmetic Progression (A.P.)

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General Term or nth Term of and Sum of an Arithmetic Progression:

  • If ‘a’ is the first term and ‘d’ is the common difference of an A.P. then the nth term is given by

tn = a + (n – 1)d

  • If ‘a’ is the first term and ‘d’ is the common difference of an A.P. then the sum of the first n terms is given by

Stn = n/2(2a + (n – 1)d) = n/2(First term + Last term)

Type – II: To Find Sum of Arithmetic Progression (A.P.).

Example – 01:

  • Find the sum of 3 + 8 + 13 + 18 + …. + n terms
  • Solution:

Given A.P. 3, 8, 13, 18 + …. + n terms

First term = a = 3, common difference = d = 8 – 3 = 5



The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Sn = n/2(2 x 3 + (n – 1)5) = n/2(6 + 5n – 5)  = n/2( 5n + 1) = n(5n+1)/2

Ans: The sum of n terms of the A.P. is n(5n+1)/2

Example – 02:

  • Find the sum of 1 + 4 + 7 + 10 + …. + 22 terms
  • Solution:

Given A.P. 1, 4, 7, 10,…..  n terms



First term = a = 1, common difference = d = 4 – 1 = 3

The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Sn = n/2(2 x 1 + (n – 1)3) = n/2(2 + 3n – 3)  = n/2( 3n – 1) = n(3n – 1)/2

∴  S22 = 22(3 x 22 – 1)/2 = 11(66 – 1) = 11 x 65 = 715

Ans: The sum of 22 terms of the A.P. is 715.



Example – 03:

  • For an A.P. a = 3, d =4. Find S20.
  • Solution:

Given first term = a = 3, common difference = d =4

The sum of n term of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Sn = n/2(2 x 3 + (n – 1)4) = n/2(6 + 4n – 4)  = n/2( 4n + 2) = n(2n + 1)

∴  S20 =  20 x (2 x 20 + 1) = 20 x 41 = 820

Ans: S20 = 820



Example – 04:

  • For an A.P. S16 = 784, a = 4. Find d.
  • Solution:

Given first term = a = 4,  S16 = 784

The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  S16 = 16/2(2 x 4 + (16 – 1)d)

∴   784 = 8(8 + 15d)

∴   784 = 64 + 120d



∴   120 d = 784 – 64 = 720

∴   d = 6

Ans: d = 6

Example – 05:

  • For an A.P. S12 = – 78, d = – 3. Find a.
  • Solution:

Given S12 = 3, common difference = d = – 3



The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  S12 = 12/2(2 a + (12 – 1) x (-3))

∴   – 78 = 6(2a – 33)

∴   – 13 = 2a – 33

∴   2a = -13 +  33 = 20

∴   a = 10



Ans: a = 10

Example – 06:

  • For an A.P. t3 = 17 and t7 =  37. Find S16.
  • Solution:

Given for an A.P. t3 = 17 and t7 =  37

The nth term of an A.P. is given by tn = a + (n – 1)d

∴  t3 = a + (3 – 1)d = 17



∴  a + 2d = 17  ………… (1)

∴  t7 = a + (7 – 1)d = 37

∴  a + 6d = 37  ………… (2)

Subtracting equation (1) from (2) we get

4 d = 30

∴  d = 5



Substituting in equation (1) we get

a + 2 x 5 = 17

∴  a + 10 = 17

∴  a = 7

The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  S16 = 16/2(2 x 7 + (16 – 1)5) = 8(14 + 75) = 8(89) = 712

Ans: S16 = 712

Example – 07:

  • For an A.P. t7 = 13 and S14 =  203. Find S8.
  • Solution:

Given for an A.P. t7 = 13

The nth term of an A.P. is given by tn = a + (n – 1)d

∴  t7 = a + (7 – 1)d = 13

∴  a + 6d = 13  ………… (1)

Given for an A.P. S14 =  203

The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  S14 = 14/2(2a + (14 – 1)d)

∴ 203 = 7(2a +13d)

∴ 2a +13d = 29   ………… (2)

Multiplying equation (1) by 2 and subtracting from equation (2)

d = 3

Substituting in equation (1) we get

a + 6 x 3 = 13

∴  a + 18 = 13

∴  a = – 5

The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  S8 = 8/2(2 x (-5) + (8 – 1)3) = 4(- 10 + 21) = 4(11) = 44

Ans: S8 = 44

Example – 08:

  • Find n if 1 + 4 + 7 + 10 + …… +n terms = 590
  • Solution:

Given A.P. 1, 4, 7, 10,…..  n terms

First term = a = 1, common difference = d = 4 – 1 = 3

The sum of n term of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Sn = n/2(2 x 1 + (n – 1)3) = n/2(2 + 3n – 3)  = n/2( 3n – 1) = n(3n – 1)/2

∴  590 = n(3n – 1)/2

∴  1180 = 3n2 – n

∴  3n2 – n – 1180 = 0

∴  3n2 – 60n + 59 n – 1180 = 0

∴  3n(n – 20) + 59(n – 20) = 0

∴  (n – 20) (3n + 59) = 0

∴  (n – 20) = 0 or (3n + 59) = 0

∴  n = 20 or n = -59/3

Now n ∈ N, hence n = -59/3 is not possible

Ans: n = 20

Example – 09:

  • Find n if 50 + 46 + 42 + 38 + …… +n terms = 336
  • Solution:

Given A.P. 50, 46, 42, 38,…..  n terms

First term = a = 50, common difference = d = 46 – 50 = – 4

The sum of n term of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Sn = n/2(2 x 50 + (n – 1)(-4)) = n/2(100 – 4n + 4)  = n/2( 104 – 4n) = n(52 – 2n)

∴  336 = n(52 – 2n)

∴  336 = 52n – 2n2

∴  2n2 – 52n + 336 = 0

∴  2n2 – 28n – 24n + 336 = 0

∴  2n(n – 14) –  24(n – 14) = 0

∴ (n – 14)(2n – 24) = 0

∴ (n – 14) = 0 or (2n – 24) = 0

∴  n = 14 or n =12

Ans: n = 14 or n = 12

Example – 10:

  • Find n if 25 + 22+ 19+ 16 + …… +n terms = 116
  • Solution:

Given A.P. 25, 22, 19, 16,…..  n terms

First term = a = 25, common difference = d = 22 – 25 = -3

The sum of n term of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Sn = n/2(2 x 25 + (n – 1)(-3)) = n/2(50 – 3n + 3)  = n/2( 53 – 3n) = n(53 – 3n)/2

∴  116 = n(53 – 3n)/2

∴  232 = n(53 – 3n)

∴  232 = 53n – 3n2

∴  3n2 – 53n + 232 = 0

∴  3n2 – 24n – 29 n – 232 = 0

∴  3n(n – 8) –  29(n – 8) = 0

∴ (n – 8)(3n – 29) = 0

∴ (n – 8) = 0 or (3n – 29) = 0

∴  n = 8 or n = 29/3

Now n ∈ N, hence n = 59/3 is not possible

Ans: n = 8

Example – 11:

  • Find the sum of all natural numbers from 1 t0 200 which is divisible by 5.
  • Solution:

Given numbers are 1, 2, 3, ……., 198, 199, 200

In forwward pass through given set of numbers, the first number divisible by 5 is 5

In backward pass through given set of numbers, the first number divisible by 5 is 200

Thus the required sequence is 5, 10, 15, 20, ……, 200

First term = a = 5, common difference = 10 – 5 = 5, tn = 200

The nth term of an A.P. is given by tn = a + (n – 1)d

∴ 200 = 5 + (n – 1)5

∴ 200 = 5 + 5n – 5

∴ 200 = 5n

∴  n = 40

The sum of n term of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Required sum = Sn = 40/2(2 x 5 + (40 – 1)(5)) = 20(10 + 195) = 20 x 205 = 4100

Ans: The sum of all natural numbers from 1 t0 200 which is divisible by 5 is 4100.

Example – 12:

  • Find the sum of all natural numbers from 1 t0 200 which is divisible by 3.
  • Solution:

Given numbers are 1, 2, 3, ……., 198, 199, 200

In forwward pass through given set of numbers, the first number divisible by 3 is 3

In backward pass through given set of numbers, the first number divisible by 3 is 198

Thus the required sequence is 3, 6, 9,12, …., 198

First term = a = 3, common difference = 6 – 3 = 3, tn = 198

The nth term of an A.P. is given by tn = a + (n – 1)d

∴ 198 = 3 + (n – 1)3

∴ 198 = 3 + 3n – 3

∴ 198 = 3n

∴  n = 66

The sum of n terms of an A.P. is given by Sn = n/2(2a + (n – 1)d)

∴  Required sum = Sn = 66/2(2 x 3 + (66 – 1)(3)) = 33(6 + 195) = 33 x 201 = 6633

Ans: The sum of all natural numbers from 1 t0 200 which is divisible by 3 is 6633.

Science > Mathematics > Sequence and SeriesYou are Here
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