Question

Given a sequence of 4 numbers, first three of which are in A.P. and the last three in G.P. with common ratio $\frac{3}{2}$.If 1 is subtracted from the last number, all the four numbers in order form an A.P., then the value of sum of smallest and largest of the four numbers is

• A. 10
• B. 13
• C. 12
• D. 11

Answer 1

Answer: (D)

4 numbers:
$a-d,a,a+d,a+2d+1$
$\frac{3}{2}a=(a+d),$
$(a+2d+1)=(a+d)\cdot \frac{3}{2}$
$3a=2a+2d$
$2a+4d+2=3a+3d$
$a=2d\dots ...(i)$
$a-d=2$ .....(ii)
From (i) and (ii),
$a=4,d=2$
$\therefore$ smallest number $=2$, largest number $=9$