Question

Find the sum of the first 25 terms of an A.P. whose nth term is given by $a_{n}$ = 7 - 3n.

Answer 1

Answer:

The sum of first 25 terms of an A.P. is - 800.

Step-by-step explanation:

Given :  

nth term, an = 7 – 3n……………(1)

On putting n = 1 in eq 1,

a1 = 7– 3(1)  

a1 = 7 - 3

a1 = 4

 

On putting n = 25 in eq 1,

a25 = 7 – 3(25)  

a25 = 7 - 75

a25 = l (last term) = - 68

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

S25 = 25/2 (4 - 68)

S25 = 25/2 × - 64

S25 = 25 × - 32  

S25 = - 800

Hence, the sum of first 25 terms of an A.P. is - 800.

HOPE THIS ANSWER WILL HELP YOU….

 

Answer 2

Answer:

-800

Explanation:

Given

nth term = $a_{n}$ = 7 - 3n.

To Find:

Sum of first 25 terms of A.P.

Solution

The nth term is,

$a_{n}$ = 7 - 3n.

First term

$a_{1}$ = 7 - 3(1)

$a_{1}$ = 7 - 3

$a_{1}$ = 4

second term

$a_{2}$ = 7 - 3(2)

$a_{2}$ = 7 - 6

$a_{2}$ = 1

Third term

$a_{3}$ = 7 - 3(3)

$a_{3}$ = 7 - 9

$a_{3}$ = -2

The series of A.P. is as follows

4, 1 , -2 ......

we have

a = 4

d = 1 - 4 = -3

n = 25

By the identity

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

The sum of 25 terms is

$S_{25} = \dfrac{25}{2}[2(4)+(25-1)(-3)]$

$S_{25} = \dfrac{25}{2}[8+(24)(-3)]$

$S_{25} = \dfrac{25}{2}[8-72]$

$S_{25} = \dfrac{25}{2}(-64)$

$S_{25} = 25(-32)$

$S_{25} = -800$