Question

There are two A.P.s whose common differences differ by unity, but sum of the three consecutive terms in each is $15$. If P and $P_1$ be the products of these terms such that $\frac{P}{P_1}=\frac{7}{8}$, then find the two A.P.'s.

Answer 1

$a-d,a,a+d$
and $A-p,A,A+p$ where $p=d+1$ ..$\left(1\right)$
$S=3a=3A=15$
$\therefore a=A=5$
Also $\frac{P}{P_1}=\frac{a(a^2-d^2)}{A(A^2-p^2)}=\frac{7}{8}$
$8(25-d^2)=7[25-(d+1{)}^2]$, by $\left(1\right)$
or $200-175=8d^2-7(d^2+2d+1)$.
$\therefore d^2-14d-32=0$ or $d=-2,16$
and hence $p=d+1=-1,17$. Also $a=A=5$
Hence the two A.P.'s are given by $7,5,3$_________ and $6,5,4$________ or $-11,5,21$, ______ and $-12,5,22$________ whose three consecutive terms written above satisfy the given conditions.