JEE PYQ: Sequence And Series Question 54
Question 54 - 2019 (10 Apr Shift 2)
Let $a, b$ and $c$ be in G.P. with common ratio $r$, where $a \neq 0$ and $0 < r \leq \frac{1}{2}$. If $3a, 7b$ and $15c$ are the first three terms of an A.P., then the $4^{th}$ term of this A.P. is:
(1) $\frac{2}{3}a$
(2) $5a$
(3) $\frac{7}{3}a$
(4) $a$
Show Answer
Answer: (4)
Solution
$\therefore a, b, c$ are in G.P. $\Rightarrow b = ar, c = ar^2$
$\therefore 3a, 7b, 15c$ are in A.P. $\Rightarrow 3a, 7ar, 15ar^2$ are in A.P.
$\therefore 14ar = 3a + 15ar^2$
$\Rightarrow 15r^2 - 14r + 3 = 0 \Rightarrow r = \frac{1}{3}$ or $\frac{3}{5}$ rejected
$\therefore r < \frac{1}{2} \therefore r = \frac{1}{3}$
Fourth term $= 15ar^2 + 7ar - 3a$
$= a\left(15 \cdot \frac{1}{9} + 7 \cdot \frac{1}{3} - 3\right) = a$
BETA
