Question

There are two sets of numbers each consisting of $3$ terms in A.P. and the sum of each set of numbers is $12$. The common difference of the first set is greater by $1$ than the common difference of the second set. The product of the first set of numbers and the product of the second set of numbers is in the ratio $5:4$. Find the numbers in both the sets.

Answer 1

Let the first set be, $a-d,a,a+d$

Given, $a-d+a+a+d=12\Rightarrow a=4$

$\Rightarrow 4-d,4,4+d$........(1)

Let the second set be, $a-d+1,a,a+d-1$

$a=4\Rightarrow 5-d,4,3+d$..........(2)

$\left(1\right)÷\left(2\right)$ gives,

$\frac{(4-d)\times 4\times (4+d)}{(5-d)\times 4\times (3+d)}=\frac{5}{4}$

$\frac{d^2-16}{d^2-2d-15}=\frac{5}{4}$

$4d^2-64=5d^2-10d-75$

$d^2-10d-11=0$

$d=11,-1$

when, $d=11$

first set $a-d,a,a+d=4-11,4,4+11$

$=-7,4,15$

second set $a-d+1,a,a+d-1=4-11+1,4,4+11-1$

$=-6,4,14$

when, $d=-1$

first set $a-d,a,a+d=4-(-1),4,4-1$

$=5,4,3$

second set $a-d+1,a,a+d-1=4-(-1)+1,4,4+(-1)+1$

$=6,4,2$