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The sum of four consecutive numbers in an A.P. is $32$ and the ratio of the product of the first and the last term to the product of two middle terms is $7 : 15$. Find the numbers.
Given: Sum of four consecutive number of A.P. $=32$ and the ratio of the product of the first and the last term to the product of two middle terms $=7 : 15$
To do: Find the numbers.
Solution:
Let the numbers be $( a–3d),\ ( a–d),\ ( a+d)\ and\ ( a+3d)$
Sum of the numbers $=a-3d+a-d+a+d+a+3d=32$
$\Rightarrow 4a=32$
$\Rightarrow a=\frac{32}{4}=8$
Product of first and last numbers $=(a-3d)(a+3d)=a^{2}-9d^{2}$
Product of the two middle numbers $=(a-d)(a+d)=a^{2}-d^{2}$
As given,
$\frac{a^{2}-9d^{2}}{a^{2}-d^{2}}=\frac{7}{15}$
$\Rightarrow 15(a^{2}-9d^{2})=7( a^{2}-d^{2})$
$\Rightarrow 15a^{2}-135d^{2}=7a^{2}-7d^{2}$
$\Rightarrow 15a^{2}-7a^{2}=135d^{2}-7d^{2}$
$\Rightarrow 8a^{2}=128d^{2}$
$\Rightarrow d^{2}=\frac{8a^{2}}{128}=\frac{a^{2}}{16}$
$\Rightarrow d=±\sqrt{\frac{a^{2}}{16}}$
$\Rightarrow d=±\frac{a}{4}$
$\Rightarrow d=±\frac{8}{4}=±2$
If $d=2$ numbers are:
$2,\ 6,\ 10,\ 14$
If $d=−2$ numbers are:
$14,\ 10,\ 16,\ 2$
Updated on: 10-Oct-2022
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