Q. Three numbers whose sum is 15 are in AP. If they are added by 1, 4 and 19 respectively they are in GP. The numbers are
A) $2,\ 5,\ 8$
B) $26,\ 5,\ - 16$
C) $2,\ 5,\ 8$ and $26,\ 5,\ - 16$
D) None of these
Explanation
(c)
Let the three numbers in AP are $a - d,a,a + d$
Since, $a - d + a + a + d = 15\ \Rightarrow a = 5$
Since, $(a - d + 1),(a + 4),(a + d + 19)$ are in GP.
$\therefore(a + 4)^{2} = (a - d + 1)(a + d + 19)$
$\Rightarrow \ 9^{2} = (6 - d)(24 + d)\ \ \ \lbrack\because a = 5\rbrack$
$\Rightarrow \ d^{2} + 18d - 63 = 0$
$\Rightarrow \ d = 3, - 21$
$\therefore$ Required series are 2, 5, 8 and 26, 5, -16
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.