Question

There are two sets of numbers each consisting of $3$ terms in A.P. and the sum of each set is $15$. The common difference of the first set is greater by $1$ than the common difference of the second set, and the ratio of product of elements of the first set to that of the second set is $7:8$. Find the numbers.

Answer 1

Let the numbers in the first set be $(a-d),a,(a+d)$

The numbers in the second set be $A-D,A$ and $A+D$

where $d=D+1$

Now, $(a-d)+a+(a+d)=15$ and $(A-D)+A+(A+D)=15$

$⟹a=5$ and $A=5$

Also,

$\frac{(a-d)a(a+d)}{(A-D)A(A+D)}=\frac{7}{8}$

$⟹\frac{a(a^2-d^2)}{A(A^2-D^2)}=\frac{7}{8}$

$⟹\frac{5^2-d^2}{5^2-D^2}=\frac{7}{8}$ $[\because a=A=5]$

$⟹8(25-d^2)=7(25-D^2)$

$⟹200-8d^2=175-7D^2$

$8d^2-7D^2=25$

$8{(D+1)}^2-7D^2=25$ $\because [d=D+1]$

$D^2+16D-17=0$

$(D+7D)(D-1)$

$⟹D=1$ $[D\ne -17]$

$\therefore d=2$

$\therefore$ On Set $I:$ $3,5,7$

$\therefore$ On Set $II:$ $4,5,6$