Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and $q$ respectively. Let d and D be the common differences of $\mathrm{AP}^{\prime} \mathrm{s}$ in $A$ and $B$ respectively such that $D=d+3, d>0$. If $\frac{p+q}{p-q}=\frac{19}{5}$, then $\mathrm{p}-\mathrm{q}$ is equal to
Solution
<p>Let the terms in $A$ be $a_1-d, a_1, a_1+d$</p>
<p>and in $B$ be $a_2-D, a_2, a_2+D$</p>
<p>Now $3 a_1=36$</p>
<p>$\Rightarrow \quad a_1=12$</p>
<p>and $3 a_2=36$</p>
<p>$\Rightarrow \quad a_2=12$</p>
<p>Now $(12-d)(12)(12+d)=p$</p>
<p>and $(12-D)(12)(12+D)=q$</p>
<p>Also $\frac{p+q}{p-q}=\frac{19}{5}$</p>
<p>$$\begin{aligned}
& \Rightarrow 12 q=7 p \\
& \Rightarrow 12(12-D)(12)(12+D)=7(12-d)(12)(12+\mathrm{d}) \\
& \Rightarrow 12(9-d)(12)(15-d)=7(12-d)(12)(12+\mathrm{d}) \\
& \Rightarrow 12\left(135-d^2-6 d\right)=7\left(144-d^2\right) \\
& \Rightarrow d=6, D=9 \\
& p=6 \times 12 \times 18=1296 \\
& q=756 \\
& p-q=540
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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