Question

find the common difference if an AP whose forst term is 5 and the sum of first four terms is half of the sum of next four terms. ​

Answer 1

Answer:

••hey mate ••

••here is the answer ••

Let d is common difference of AP

Let d is common difference of APnow first 4term is 5,5+d,5+2d,5+3d

Let d is common difference of APnow first 4term is 5,5+d,5+2d,5+3dand next 4term 5+4d,5+5d,5+6d,5+7d

Let d is common difference of APnow first 4term is 5,5+d,5+2d,5+3dand next 4term 5+4d,5+5d,5+6d,5+7dnow according to question

Let d is common difference of APnow first 4term is 5,5+d,5+2d,5+3dand next 4term 5+4d,5+5d,5+6d,5+7dnow according to question20+6d=(20+22d)/2

Let d is common difference of APnow first 4term is 5,5+d,5+2d,5+3dand next 4term 5+4d,5+5d,5+6d,5+7dnow according to question20+6d=(20+22d)/220+6d=10+11d

Let d is common difference of APnow first 4term is 5,5+d,5+2d,5+3dand next 4term 5+4d,5+5d,5+6d,5+7dnow according to question20+6d=(20+22d)/220+6d=10+11dd=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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