Consider two sets A and B, each containing three numbers in A.P.

Question

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

Accepted Answer

Let the terms in $A$ be $a_1-d, a_1, a_1+d$

and in $B$ be $a_2-D, a_2, a_2+D$

Now $3 a_1=36$

$$\Rightarrow \quad a_1=12$$

and $3 a_2=36$

$$\Rightarrow \quad a_2=12$$

Now $(12-d)(12)(12+d)=p$

and $(12-D)(12)(12+D)=q$

Also $\frac{p+q}{p-q}=\frac{19}{5}$

$$\begin{aligned}
& \Rightarrow 12 q=7 p \
& \Rightarrow 12(12-D)(12)(12+D)=7(12-d)(12)(12+\mathrm{d}) \
& \Rightarrow 12(9-d)(12)(15-d)=7(12-d)(12)(12+\mathrm{d}) \
& \Rightarrow 12\left(135-d^2-6 d\right)=7\left(144-d^2\right) \
& \Rightarrow d=6, D=9 \
& p=6 \times 12 \times 18=1296 \
& q=756 \
& p-q=540
\end{aligned}$$

Solutions · 1