Question

A sequence of three real numbers form an A.P. and the first term is 9. If 2 is added to the second term and 20 is added to the third term, the resulting numbers will form a G.P. The least value of the third term of the G.P. is

• A. 49
• B. 1
• C. - 3
• D. - 19

Answer 1

Answer: (B)

$9,9+d,9+2d\text{ in A.P. }$
Also $9,11+d,29+2d$ in G.P.
Hence $(11+d{)}^2=9(29+2d)$
Simplifying$d^2+4d-140=0$
$(d+14)(d-10)=0$
$d=-14$ or $d=10$
$\Rightarrow$ possible A.P.'s are $9,-5,-19\dots$ or $9,19,29$
Possible G.P. is $9,-3,1$ or $9,21,49$ with common ratio $-1/3$ or $7/3$
Hence least $3^{\text{rd }}$ term is 1