Question

The sum of each of two sets of three terms in A.P. is 15. The common difference of the first set is greater than that of the second by 1 and the ratio of the products of the terms in the first set and that of the second set is 7:8 the ratio of the smallest terms in two sets of terms is

• A. $\frac{3}{4}or\frac{11}{12}$
• B. $\frac{2}{3}or\frac{1}{2}$
• C. $\frac{2}{3}or\frac{3}{4}$
• D. None

Answer 1

The correct option is A

$\frac{3}{4}or\frac{11}{12}$

Let the first set of number be $a-d,a,a+d.$
Then $a-d+a+a+d=15\Rightarrow a=5$
The second set of numbers will be
$b-(d-1),b,b+(d-1).Again,b-(d-1)+b+b+(d-1)=15\Rightarrow b=5$
Hence the sets of numbers are
$5-d,5,5+dand6-d,5,4+d.$
Further, from the given condition
$\frac{(5-d)5(5+d)}{(6-d)5(4+d)}=\frac{7}{8}\Rightarrow \frac{25-d^2}{24+2d-d^2}=\frac{1}{8}$
$\Rightarrow d^2+14d-32=0\Rightarrow d=2,-16$
$\therefore thetwosetsare3,5,7and4,5,6or21,5,-11and22,5,-12.$
$\therefore$ Ratio of their smallest term is
$\frac{3}{4}or\frac{-11}{-12}i.e.\frac{3}{4}or\frac{11}{12}$