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 next fourth term​

Answer 1

Step-by-step explanation:

Let the common difference be d.

The sum of the first four terms is

5 + (5+d) + (5+d+d) + (5+d+d+d)

= 20 + 6d

The sum of the next 4 terms is

5+4d + 5+5d + 5+6d + 5+7d = 20 + 22d

You have been told that

20 +6d = (1/2)(20 + 22d)= 10 + 11d

Therefore

5d = 10

d = 2

Check: does

5+7+9+11 = (1/2)(13+15+17+19) ?

32 = (1/2)(64)

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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