Question

Find the common difference of an AP , whose first term is 5 and sum of its first four term is half the sum of the next four terms.

Answer 1

Answer:

2

Step-by-step explanation:

$a = 5 \\ a1 + a2 + a3 + a4 = \frac{1}{2} (a5 + a6 + a7 + a8)$

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

$4a + 6d = \frac{1}{2} (4a + 22d)$

$4a + 6d = 2a + 11d$

$2a = 5d$

$d = \frac{2}{5} a$

$d = \frac{2}{5} (5)$

$d = 2$

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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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