Question

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

Answer 1

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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

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

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

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