Question

The numbers $a, b, c$ in order are three consecutive terms of an A.P. such that $a+b+c=60$. If $(a-2), b,(c+3)$ in order are the three consecutive terms of a G.P., then possible value $(s)$ of $\left(a^2+b^2\right.$ $\left.+ c ^2\right)$ is/are

• A. 1218
• B. 1208
• C. 1288
• D. 1298

Answer 1

Answer: (B)

Given $ 2 b = a + c$
$\text { and } a+b+c=60 $
$3 b =60 \Rightarrow b =20 $
$\therefore c =40- a $
$\text { now } a-2, b, c+3 \text { in G.P. } $
$a -2,20,43- a \text { in G.P. }$
$(a-2)(43-a)=400 \Rightarrow 45 a-86-a^2=400 \Rightarrow a^2-45 a+486=0 $
$a^2-27 a-18 a+486=0 \Rightarrow a(a-27)-18(a-27)=0 $
$a =27 \text { or } a =18 $
$a =27, c =13 $
$a =18, c =22 $
$\therefore 27,20,13 \text { or } 18,20,22$
$a ^2+ b ^2+ c ^2=729+400+169=1298 \Rightarrow \text { (D) }$
$a^2+b^2+c^2=324+400+484=1208 \Rightarrow \text { (B) }$