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 product of the elements of $A$ be $36$ and $p$ respectively, and the sum and product of elements of $B$ be $36$ and $q$ respectively. Let $d, D$ be the common differences of $A, B$ respectively with $D=d+3$, $d>0$. If $\dfrac{p+q}{p-q}=\dfrac{19}{5}$, then $p-q$ is equal to:

Accepted Answer

Write $A = \{12-d,12,12+d\}$, $B=\{12-D,12,12+D\}$ (each has middle term $12$ since sum$=36$).

$p=(12-d)(12)(12+d) = 12(144-d^2)$, $q = 12(144-D^2)$.

From $\dfrac{p+q}{p-q}=\dfrac{19}{5}$: $5(p+q)=19(p-q) \Rightarrow 12q=7p$.

Substituting $D=d+3$ and solving the resulting equation gives $d=6$, $D=9$.

$$p = 12(144-36)=1296, \qquad q = 12(144-81) = 756.$$

$p-q = 1296-756 = 540.$

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