# FIND THE SUM OF N TERMS FROM THE GIVEN SUM OF SERIES

Find the Sum of n Terms from the Given Sum of Series ?

From the sum of n terms of the series, we may form a equation and solve for the variables "a" and "d".

To find the sum of n terms of an arithmetic progression, we may use one of the formulas given below.

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

a = first term, d = common difference and n = number of terms.

## Find the Sum of n Terms from the Given Sum of Series - 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 :

Given that :

S7  =  49

Sn  =  (n/2) [2a + (n - 1)d]

S7  =  (7/2)[2a + (7-1)d]

(7/2)[2a + 6d]  =  49

2a + 6d  =  49⋅(2/7)

2a + 6d  =  14

Dividing both sides by 2

a + 3d  =  7  ------(1)

S17  =  289

S17  =  (17/2)[2a + (17-1)d]

(17/2)[2a + 16d]  =  289

2a + 16d  =  289⋅(2/17)

2a + 16d  =  34

Dividing both sides by 2

a + 8d  =  17  ------(2)

(1) - (2)

a + 3d  =  7

a + 8d  =  17

(-)   (-)   (-)

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

-5d  =  -10

d  =  2

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

a + 3(2)  = 7

a + 6  =  7

a  =  7 - 6  =  1

By applying the values of "a" and "d" in the formula for sum of n terms.

Sn  =  (n/2) [2a + (n - 1)d]

Sn  =  (n/2) [2(1) + (n - 1)2]

Sn  =  (n/2) [2 + 2n - 2]

Sn  =  n2

## From the General Term Find the Sum of Series

Question 2 :

Show that a1a2,............ an form an AP where an is defined as below

(i) an  =  3 + 4n

(ii) an  =  9 - 5n

Also find the sum of 15 terms in each case.

Solution :

(i) an  =  3 + 4n

From the given general term, we may find the first term and common difference.

By applying n = 1, we get the first term.

 n  =  1a1  =  3 + 4(1)  =  7 n  =  15a2  =  3 + 4(15)  =  63

First term (a)  =  7, Last term (l)  =  63

Number of terms  =  15

Sn  =  (n/2) [a + l]

S15  =  (15/2) [7 + 63]

=  (15/2) (70)

=  15(35)

=  525

Hence the sum of 15 terms is 525.

(ii) an  =  9 - 5n

From the given general term, we may find the first term and common difference.

By applying n = 1, we get the first term.

 n  =  1a1  =  9 - 5(1)  =  4 n  =  15a2  =  9 - 5(15)  = =  9 - 75=  -66

First term (a)  =  4,

Last term (l)  =  -66

Number of terms  =  15

Sn  =  (n/2) [a + l]

S15  =  (15/2) [4 - 66]

=  (15/2) [-62]

=  15 (-31)

=  -465

Hence the sum of 15 terms is -465. After having gone through the stuff given above, we hope that the students would have understood, find the sum of n terms from the given sum of series.

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