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 = 1

t_{n} = 4n - 3

t_{1} = 4(1) - 3 = 4 - 3

t_{n} = 1 ==> a = 1

n = 2

t_{n} = 4n - 3

t_{2} = 4(2) - 3 = 8 - 3

t_{2} = 5

d = 5 - 1 = 4

S_{n} = (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 2n^{2} -3n . Show that the series is an A.P.

Solution :

Given that :

t_{n} = 2n^{2} - 3n

t_{n} = 2n^{2} - 3n

n = 1

t_{1} = 2(1)^{2} - 3(1)

= 2 - 3

t_{1 } = -1 = a

n = 2

t_{2} = 2(2)^{2} - 3(2)

= 8 - 6

t_{2 } = 2

d = 2 - (-1) = 3

S_{n} = (n/2) [2a + (n - 1)d]

S_{n} = (n/2) [2(-1) + (n - 1)3]

= (n/2) [-2 + 3n - 3]

= (n/2) [3n - 5]

Question 3 :

The 104^{th} term and 4^{th} 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

S_{n} = (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]

S_{35} = 612.5

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