JEE PYQ: Sequence And Series Question 33
Question 33 - 2020 (06 Sep Shift 1)
Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) = 0$. Then:
(1) $a, c, p$ are in A.P.
(2) $a, c, p$ are in G.P.
(3) $a, b, c, d$ are in G.P.
(4) $a, b, c, d$ are in A.P.
Show Answer
Answer: (3)
Solution
Rearrange given equation, we get
$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) = 0$
$\Rightarrow (ap - b)^2 + (bp - c)^2 + (cp - d)^2 = 0$
$\therefore ap - b = bp - c = cp - d = 0$
$\Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$ $\therefore a, b, c, d$ are in G.P.
BETA
