Question

Find the common difference of an ap whose first term is 5 and the sum of its first 4 terms is half of its next four terms

Answer 1

a=5
sum of its first 4 terms =
a + (a+d) + (a+2d) + (a+3d)

sum of its next 4 terms =
(a+4d) + (a+5d) + (a+6d) + (a+7d)

Since, the sum of its first 4 terms is half of its next four terms
So,
[ a + (a+d) + (a+2d) + (a+3d) ] * 2 = (a+4d) + (a+5d) + (a+6d) + (a+7d)
(4a + 6d) * 2 = 4a + 22d
8a + 12d = 4a + 22d
4a = 10d
2a = 5d
Now putting value of a
2 * 5 = 5 * d
Therefore,
d = 2

Answer 2

$let \: first \: term \: be \: \: a \: and \: common \\ difference \: be \: \: d. \\ given \: a = 5 \: \: and \: \\ sum(t1 \: to \: t4) = \frac{1}{2} sum(t5 \: to \: t8) - - (1)$
$we \: know \\ sum(t1 \: to \: tn) = \frac{n}{2} \times (t1 + tn)$
$thus \: (1) \: becomes \\ \frac{4}{2} \times (a + (a + 3d)) = \frac{1}{2} \times \frac{4}{2} \times ((a + 4d) + (a + 7d)) \\ i.e. \: 4a + 6d = 2a + 11d \\ d = \frac{2}{5} a = 2$