Question

find the common difference of an AP whose first term is 1 and the sum of the first four terms is one-third to the sum of the next four term.

Answer 1

hope this will help u

Answer 2

Answer:

$common\: difference (d)=2$

Step-by-step explanation:

Given first term (a) = 1,

Let the common difference = d,

According to the problem given,

The sum of the first four terms is one-third to the sum of the next four term.

$1+(1+d)+(1+2d)+(1+3d)<strong>=</strong>\frac{1}{3}[(1+4d)+(1+5d)+(1+6d)+(1+7d)]$

$\implies 4+6d=\frac{1}{3}(4+22d)$

$\implies 3(4+6d)=4+22d$

$\implies 12+18d=4+22d$

$\implies 12-4=22d-18d$

$\implies 8 = 4d$

$\implies \frac{8}{4}=d$

$\implies 2=d$

Therefore,

$common\: difference (d)=2$

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