Question

find the common difference of AP whose first term is 5 and sum of first four terms is half the sum of next 4terms

Answer 1

a=5, S 4= 1/2(S 8 - S4)
n/2(2a+(n-1)d)=1/2(n/2(2a+(n-1)d) -n/2(2a+(n-1)d)
2(2×5+3d)=1/2(4(2×5+7d) -2(2×5+3d))
20+6d=1/2(40+28d - 20-6d)
20+6d= 1/2(20+22d)
20+6d=10+11d
20-10=11d-6d
10=5d
d=10/5
d=2.ANS

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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