Question

Find the sum of the first 25 terms of an A.P. whose nth term is given by $a_{n}$ = 2 - 3n.

Answer 1

Answer:

The sum of first 25 terms of an A.P. is - 925.

Step-by-step explanation:

Given :  

nth term, an = 2 – 3n……………(1)

On putting n = 1 in eq 1,

a1 = 2 – 3(1)  

a1 = 2 - 3

a1 = -1

 

On putting n = 25 in eq 1,

a25 = 2 – 3(25)  

a25 = 2 - 75

a25 = l (last term) = - 73

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

S25 = 25/2 (-1 - 73)

S25 = 25/2 × - 74

S25 = 25 × - 37  

S25 = - 925

Hence, the sum of first 25 terms of an A.P. is - 925.

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Answer 2

Answer:

-925

Explanation:

Given

nth term = $a_{n}$ = 2 - 3n.

To Find:

Sum of first 25 terms of A.P.

Solution:

The nth term is,

$a_{n}$ = 2 - 3n.

First term

$a_{1}$ = 2 - 3(1)

$a_{1}$ = 2 - 3

$a_{1}$ = -1

second term

$a_{2}$ = 2 - 3(2)

$a_{2}$ = 2 - 6

$a_{2}$ = -4

Third term

$a_{3}$ = 2 - 3(3)

$a_{3}$ = 2 - 9

$a_{</strong><strong>3</strong><strong>}$ = -7

The series of A.P. is as follows

-1, -4 , -7 ......

We have

a = -1

d = -4+1 = -3

n = 25

By the identity

$\boxed{ S_{n}= \dfrac{n}{2}[2a+(n-1)d] }$

The sum of 25 terms is

$S_{25} = \dfrac{25}{2}[2(-1)+(25-1)(-3)]$

$S_{25} = \dfrac{25}{2}[-2+(24)(-3)]$

$S_{25} = \dfrac{25}{2}[-2-72]$

$S_{25} = \dfrac{25}{2}(-74)$

$S_{25} = 25(-37)$

$S_{25} = -925$