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 common term for all such possible A.P's?

Answer 1

Let $a-d,a,a+d$ and $A-p,A,A-p$ are the A.Ps where $p=d+1...\left(1\right)$
$\Rightarrow S=3a=3A=15$
$\therefore a=A=5$
Also $\frac{P}{P_1}=\frac{a(a^2-d^2)}{A(A^2-p^2)}=\frac{7}{8}$
$\Rightarrow 8[25-d^2]=7[25-{(d+1)}^2],by\left(1\right)$
$or200-175=8d^2-7[d^2+2d+1]$
$\therefore d^2-14d-32=0$
or $d=-2,16$
and hence $p=d+1=-1,17$. Also $a=A=5$
Hence the two A.P.'s
are given by $7,5,\mathrm{3...}$ and $6,5,\mathrm{4...}$ or $...-11,5,\mathrm{21...}-12,5,\mathrm{22...}$
whose three consecutive terms written above the given conditions.
Hence common for all A.Ps are $5$.