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:-

Let the common difference of the AP be d.

First term, a = 5

and now

Given:

$\sf{a_{1}+a_{2} +a_{3} +a_{4} =\frac{1}{2}(a_{5} +a_{6} +a_{7}+a_{8} })$

⇒ a + (a + d) + (a + 2d) + (a + 3d)

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

$\sf{[a_{n} =a(n-1)d]}$

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

⇒8a + 12d = 4a + 22d

⇒ 22d - 12d = 8a - 4a

⇒ 10d = 4a

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

$=\sf{d=\frac{2}{5}x5=2 }$

$\fbox{Hence, the common difference of the AP is 2.}$

Answer 2

Let the common difference of the AP be d.

First term, a = 5

and now

Given:

$</p><p>\sf{a_{1}+a_{2} +a_{3} +a_{4} =\frac{1}{2}(a_{5} +a_{6} +a_{7}+a_{8} })$

a + (a + d) + (a + 2d) + (a + 3d)

[(a+4d)+(a+5d)+(a+6d)+(a+7d)]

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