Question

Three numbers whose sum is $45$ are in A.P. If $5$ is subtracted from the first number and $25$ is added to third number, the numbers are in G.P. Then numbers can be

• A. $10,15,20$
• B. $8,15,22$
• C. $5,15,25$
• D. $12,15,18$

Answer 1

Answer: (A)

$10,15,20$

Let the numbers be $a-d,a,a+d$

Their sum is $45a-d+d+a+d=453a=45a=15$

The changed numbers are $15-d-5,15,15+d+2510-d,15,40+d$

Condition to be in GP is $b^2=ac{15}^2=(10-d)(40+d)225=400-30d-d^2{d}^2+30d-175=0d^2+35d-5d-175=0d(d+35)-5(d+35)=0(d-5)(d+35)=0d=5,-35$

$a=15,d=5$

So the series is $10,15,20$