Question

if the sum of first 8 terms of arithmetic progression is 136 and that of first 15 terms is 465, then find the sum of first 25 terms.​

Answer 1

Answer:

S25=1275

Step-by-step explanation:

Let a is first term and d is cd

So S8=8/2 { 2a+7d}=136

4 { 2a+7d}=136

2a+7d=34...................(1)

S15=15/2 { 2a+14d}=465

2a+14d=465*2/15=31*2=62

2a+14d=62...............(2)

subtract (1) from(2)

7d=28, d=4

from(1)  2a+7*4=34

2a=34-28=6

a=3

Thus sum of 25 terms

S25=25/2 ( 2*3+24*4)

=25/2( 6+96)

=25/2*102

=25*51

S25=1275

Answer 2

$\huge\boxed{\underline{\bf\green{A} \: \red{N} \: \orange{S} \: \purple{W} \: \blue{E} \: \pink{R}}}$

In an A.P. sum of first n terms is given by :-

$\longrightarrow \sf S_n=\dfrac{n}{2}\bigg[2a+(n-1)d\bigg]$

where

• a= first term

• d= common difference

As per given , we have

$\longrightarrow \sf \dfrac{8}{2}\bigg[2a+(8-1)d\bigg]=136\\\\\longrightarrow \sf 2a+7d=34 \: ........(1)$

____________________________________

$\longrightarrow \sf\dfrac{15}{2}\bigg[2a+(15-1)d\bigg]=465\\\\\longrightarrow \sf2a+14d= 62 \: ......(2)$

Subtract equation(1) from (2) , we get

$\longrightarrow \sf7d=28\\\\\longrightarrow \sf d=4$

Put value of d=4 in (1) , we get

$\longrightarrow \sf2a+7(4)=34\\ \\\longrightarrow \sf 2a=6\\\\\longrightarrow \sf a=3$

Now , sum of first 25 terms will be :

$\longrightarrow \sf S_{25}=\dfrac{25}{2} \bigg[2\times3+24(4)\bigg] \\ \\ \large\longrightarrow\boxed{ \sf S_{25}=1275}$

Hence, the sum of first 25 terms is 1275