The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number.

Given: 

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is $7:15$.

To do:

We have to 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 the first and the last numbers $=(a-3d)(a+3d)=a^{2}-9d^{2}$

Product of the two middle numbers $=(a-d)(a+d)=a^{2}-d^{2}$

According to the question,

$\frac{a^{2}-9d^{2}}{a^{2}-d^{2}}=\frac{7}{15}$

$\Rightarrow 15(a^{2}-9d^{2})=7( a^{2}-d^{2})$

$\Rightarrow 15a^{2}-135d^{2}=7a^{2}-7d^{2}$

$\Rightarrow 15a^{2}-7a^{2}=135d^{2}-7d^{2}$

$\Rightarrow 8a^{2}=128d^{2}$

$\Rightarrow d^{2}=\frac{8a^{2}}{128}=\frac{a^{2}}{16}$

$\Rightarrow d=±\sqrt{\frac{a^{2}}{16}}$

$\Rightarrow d=±\frac{a}{4}$

$\Rightarrow d=±\frac{8}{4}=±2$

If $a=8, d=2$ then

$a-3d=8-3(2)=8-6=2$

$a-d=8-2=6$

$a+d=8+2=10$

$a+3d=8+3(2)=8+6=14$

If $a=8, d=-2$ then

$a-3d=8-3(-2)=8+6=14$

$a-d=8-(-2)=8+2=10$

$a+d=8+(-2)=8-2=6$

$a+3d=8+3(-2)=8-6=2$

The numbers are $2, 6, 10$ and $14$.

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Updated on: 10-Oct-2022

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