# SUM OF N TERMS OF AN ARITHMETIC PROGRESSION

Sum of n Terms of an Arithmetic Progression :

To find the sum of the series, we use the formula given below.

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

To find the first term, common difference and number of term, we use the formula given below.

an  =  a + (n - 1)d

## Sum of n Terms of an Arithmetic Progression - Examples

Question 1 :

If the sum of 7 terms of an AP is 49 and that of 17 terms is 289,find the sum of first n terms.

Solution :

S7  =  49

S17 =  289

Sn  =   (n/2) [2a + (n - 1) d]

S7  =   (7/2) [2 a + (7 - 1) d]

49  =  (7/2) [ 2a + 6d]

(49 x 2)/7  =  2 a + 6 d

14  =  2 a + 6 d

a + 3 d  =  7 -----(1)

S17  =  (17/2) [2 a + (17 - 1) d]

289 = (17/2) [ 2a + 16d]

(289 x 2)/17 = 2 a + 16 d

34 = 2 a + 16 d

a + 8 d = 17 -----(1)

(1) - (2)

a + 3 d = 7

a + 8 d = 17

(-)  (-)    (-)

-------------

- 5 d = -10

d = 2

By applying the value of d in (1), we get

a + 3 (2)  =  7

a + 6  =  7

a  =  7 - 6

a  =  1

To find the sum of first n terms, we have to apply the values of a and in the Sn formula

Sn  =  (n/2) [2a + (n - 1) d]

Sn =  (n/2) [2(1) + (n - 1) (2)]

=  (n/2) [2 + 2 n - 2]

=  (n/2) [2 n]

=  n²

Question 2 :

Show that a₁, a₂,............ an form an AP where an is defined as below

(i) an  = 3 + 4 n

(ii) an = 9 - 5 n

Also find the sum of 15 terms in each case.

Solution :

(i) a n = 3 + 4 n

 n  =  1 a₁  =  3 + 4(1) =  7 n = 2a₂  = 3 + 4(2)     = 11 d = a₂ -  a₁  =  11 - 7d  =  4

So the series will be in the form 7 + 11 + ...........

Now we need to find sum of 15 terms

S15  =  (n/2) [ 2 a + (n - 1) d]

=  (15/2) [2(7) + (15-1) 4]

=  (15/2) [14 + 14(4)]

=  (15/2) [14 + 56]

=  (15/2) [70]

=  (15 x 35)

S15  =  525

(ii) an = 9 - 5 n

Solution :

 n = 1 a1 = 9 - 5(1)  =  4 n  =  2a2  =  9 - 5(2)  =  -1 d  =  a2 -  a1     =  -1 - 4  = -5

So the series will be in the form 4 + (-1) + ...........

Now we need to find sum of 15 terms

S15  =  (n/2) [ 2 a + (n - 1) d]

= (15/2) [2(4) + (15-1) (-5)]

=  (15/2) [8 + 14(-5)]

=  (15/2) [8 - 70]

=  (15/2) [-62]

=  15 x (-31)

=  -465

After having gone through the stuff given above, we hope that the students would have understood, sum of n terms of an arithmetic progression.

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