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.


Best answer required

Answer 1

HEY THERE!!!!

Given:-

The common difference of an AP whose first term = 5

The Sum of its first four terms is half the sum of the next four terms.

Method of Solution:-

Let 'a' is the first terms of Arithmetic

Progressions and 'd' is Common Difference of Arithmetic Progressions.

According to the Question;-

a1+a2+a3+a4=1/2(a5+a6+a7+a8)

=> a+(a+d)+(a+2d)+(a+3d)=1/2(a+4d)+(a+5d)+(a+6d)+(a+7d)

=> a+a+a+a+6d=1/2(a+a+a+a+22d)

=> 2(4a+6d)=4a+22d

=> 8(5)+12d=4(5)+22d

=> 40+12d=20+22

=> 40-20=22d-12d

=> 20=10d

=> d=20/10

=> d=2

Hence, Common Difference of this Arithmetic Progression= 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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