Question

Find the common difference of an AP whose first term is 5 and the sum
of its first four terms is half the sum of the next four terms.
[CBSE 2012]​

Answer 1

Answer:

Let d is common difference of AP

Now first 4 terms are 5, 5+d, 5+2d, 5+3d

and next 4 terms 5+4d, 5+5d, 5+6d, 5+7d

Given that, the sum of its first four terms is half the sum of the next four terms.

i.e.,

5 + 5+d + 5+2d + 5+3d=25+4d + 5+5d + 5+6d + 5+7d

20+6d=2(20+22d)

20+6d=10+11d

d=2

hence the answer is 2.

Answer 2

$\huge\bf\green{\underline{\underline{Answer :}}}$

Let d be the common difference of given A.P.

First term (a)=5

$\huge\bf\green{\underline{\underline{Given :- }}}$

a1 +a2 +a3 +a4 = $\frac{1}{2}× (a5 +a6 +a7 +a8)$

⇒a+(a+d)+(a+2d)+(a+3d) = $\frac{1}{2} [(a+4d)+(a+5d)+(a+6d)+(a+7d)]$

⇒4a+6d = $\frac{1}{2} (4a+22d)$

⇒4a+6d = 2a+11d

⇒11d−6d = 4a−2a

⇒d = $\large\frac{2×5}{5}$

⇒d = 2

$\bf{Hence\: the\:common\: difference\:of\:}$

$\bf{given\: A.P. \:is \:2.}$

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