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Algebra

Sum of Arithmetic Progression

In the last article, we have studied the general term of an arithmetic progression. In this article, we shall study to find the sum of arithmetic progression.

General Term or nth term 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

Sum up to nth term of an Arithmetic Progression

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

Sum of Arithmetic Progression

Sn = n/2(First term + Last term)

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)

∴  Sn = n/2(6 + 5n – 5) 

∴  Sn = n/2( 5n + 1)

∴  Sn = 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 sum of arithmetic progression of n terms is given by Sn = n/2(2a + (n – 1)d)

∴  Sn = n/2(2 x 1 + (n – 1)3)

∴  Sn = n/2(2 + 3n – 3) 

∴  Sn = n/2( 3n – 1)

∴  Sn = 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 arithmetic progression 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)

∴  Sn = n/2(6 + 4n – 4) 

∴  Sn = n/2( 4n + 2)

∴  Sn = 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 arithmetic progression 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) 

∴  S16 = 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) 

∴  S8 = 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)

∴  Sn = n/2(2 + 3n – 3) 

∴  Sn = n/2( 3n – 1)

∴  Sn = 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))

∴  Sn = n/2(100 – 4n + 4) 

∴  Sn = n/2( 104 – 4n)

∴  Sn = 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))

∴  Sn = n/2(50 – 3n + 3) 

∴  Sn = n/2( 53 – 3n)

∴  Sn = 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))

∴  Sn = 20(10 + 195) = 20 x 205 = 4100

Ans: The sum of all natural numbers from 1 to 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))

∴  Sn = 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.

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