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

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Step 1: Let four consecutive numbers be $a - 3d, a - d, a + d, a + 3d$.
Sum $= 4a = 32 \Rightarrow a = 8$
Step 2: $\frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{7}{15}$
$\Rightarrow \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \Rightarrow 15(64 - 9d^2) = 7(64 - d^2)$
Step 3: $\Rightarrow d^2 = 4 \Rightarrow d = \pm 2$
Step 4: Four numbers are $2, 6, 10, 14$.

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