Question

The sum of four consecutive numbers in an $$AP$$ 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.

Solution

Let,the numbers in A.P be $$a-3d,a-d,a+d,a+3d$$ Given: Sum of four consecutive numbers $$=32$$ $$\therefore a-3d+a-d+a+d+a+3d=32$$ $$\implies 4a=32$$ $$\implies a=\dfrac{32}{4}=8$$ Now, A.T.Q, $$\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}=\dfrac{7}{15}$$ $$\implies \dfrac{a^2-9d^2}{a^2-d^2}=\dfrac{7}{15}$$ $$\implies \dfrac{64-9d^2}{64-d^2}=\dfrac{7}{15}[\because a=8]$$ $$\implies (64-9d^2)\times 15=7(64-d^2)$$ $$\implies 64\times 15-135d^2=7\times 64-7d^2$$ $$\implies 64\times 15-7\times 64=135d^2-7d^2$$ $$\implies 64(15-7)=128d^2$$ $$\implies 64\times8=128d^2$$ $$\implies \dfrac{64 \times8}{128}=d^2$$ $$\implies d^2=5$$ $$\implies d=\pm2$$ $$\therefore$$ Either , the numbers are $$2,6,10,14$$ or $$14,10,6,2$$Mathematics

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