Question

Divide 56 in four parts in AP such that the ratio of the product of their extremes (1^st and 4^th) to the product of means (2^nd and 3^rd) is 5 : 6 ?

Answer 1

Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d.
Given:
(a − 3d) + (a − d) + (a + d) + (a + 3d) = 56
⇒ 4a = 56
⇒ a = 14

\[Also, \]
\[\frac{\left( a - 3d \right)\left( a + 3d \right)}{\left( a - d \right)\left( a + d \right)} = \frac{5}{6}\]
\[ \Rightarrow \frac{a^2 - 9 d^2}{a^2 - d^2} = \frac{5}{6}\]
\[ \Rightarrow \frac{\left( 14 \right)^2 - 9 d^2}{\left( 14 \right)^2 - d^2} = \frac{5}{6}\]
\[ \Rightarrow \frac{196 - 9 d^2}{196 - d^2} = \frac{5}{6}\]

\[\Rightarrow 1176 - 54 d^2 = 980 - 5 d^2 \]
\[ \Rightarrow 196 = 49 d^2 \]
\[ \Rightarrow d^2 = 4\]
\[ \Rightarrow d = \pm 2\]

When d = 2, the terms of the AP are 8, 12, 16, 20. When d = −2, the terms of the AP are 20, 18, 12, 8.