# HOW TO FIND SUM OF FIRST N TERMS WHEN nth TERM IS GIVEN

How to Find Sum of First n Terms When nth Term is Given :

Here we are going to see, how to find sum of first n terms when nth term is given.

## How to Find Sum of First n Terms When nth Term is Given ?

A series whose terms are in Arithmetic progression is called Arithmetic series.

To find sum of arithmetic series, we use one of the formulas given below.

Sn  =  (n/2) [a + l]

Sn  =  (n/2) [2a + (n - 1)d]

a = first term,

n = number of terms of the series,

d = common difference and l = last term

Question 1 :

Find the sum of first 28 terms of an A.P. whose nth term is 4n -3.

Solution :

tn  =  4n - 3

n = sum of 28 terms  =  28

 n = 1tn  =  4n - 3t1  =  4(1) - 3  = 4 - 3tn  =  1  ==> a = 1 n = 2tn  =  4n - 3t2  =  4(2) - 3  = 8 - 3t2  =  5 d = 5 - 1 = 4

Sn  =  (n/2) [2a + (n - 1)d]

=  (28/2) [2(1) + (28 - 1)4]

=  14[2 + 27(4)]

=  14 [110]

=  1540

Hence the sum of 28 terms of the given sequence is 1540.

Question 2 :

The sum of first n terms of a certain series is given as 2n2 -3n . Show that the series is an A.P.

Solution :

Given that :

tn  =  2n2 - 3n

 tn  =  2n2 - 3nn = 1t1  =  2(1)2 - 3(1)  =   2 - 3   t1   =  -1  =  a n = 2t2  =  2(2)2 - 3(2)  =   8 - 6   t2   =  2d  =  2 - (-1)  =  3

Sn  =  (n/2) [2a + (n - 1)d]

Sn  =  (n/2) [2(-1) + (n - 1)3]

=  (n/2) [-2 + 3n - 3]

=  (n/2) [3n - 5]

Question 3 :

The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.

Solution :

104th term  =  125

4th term  =  0

a + 103d  = 125   -----(1)

a + 3d  =  0   -----(2)

(1) - (2)

a + 103d - a - 3d  =  125 - 0

100d  =  125

d = 125/100  =  5/4

By applying the value of d in (2), we get

a + 3(5/4)  =  0

a  =  -15/4

Sn  =  (n/2) [2a + (n - 1)d]

=  (35/2) [2(-15/4) + (35 - 1)(5/4)]

=  (35/2)[(-15/2) + (85/2)]

=  (35/2)[70/2]

S35  =  612.5

After having gone through the stuff given above, we hope that the students would have understood, "How to Find Sum of First n Terms When nth Term is Given".

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