Question

(ii) Find the common difference of an AP whose first term is 5
and the sum of the first four terms is half the sum of the next
four terms.
​.​

Answer 1

Answer:

$hence = the \: common \\ \: difference \: of \: the \: {(ap)} \: \: is \: 2$

Answer 2

Answer :

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

First term (a)=5

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

Hence the common difference of given A.P. is 2.

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