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 $(a^2+b^2$ $+c^2)$ is/are

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

Answer 1

Answer: (B)

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