JEE Main 2021 (Online) 26th February Evening ShiftQuestion 1
If the arithmetic mean and geometric mean of the $p^{\text{th}}$ and $q^{\text{th}}$ terms of the sequence $-16,8,-4,2,\ldots$ satisfy the equation $4x^{2}-9x+5=0$, then $p+q$ is equal to $\_\_\_\_\_\_$ .
Sequence and Series · Geometric Progression · Means and Inequalities (AM, GM, HM)
Answer$10$
Solution
Formulas used. (1) $t_n=ar^{n-1}$(2) $\text{G.M.}\le\text{A.M.}$ and $\text{G.M.}^2=t_pt_q$Step 1 - Roots. $4x^2-9x+5=(x-1)(4x-5)=0$, so the means are $1$ and $\tfrac54$. Since G.M. $\le$ A.M.: A.M. $=\tfrac54$, G.M. $=1$. Step 2 - Product of the terms. $t_n=-16\left(-\tfrac12\right)^{n-1}$: $$t_pt_q=256\left(-\tfrac12\right)^{p+q-2}=1\ \Longrightarrow\ 2^{p+q-2}=256=2^8 .$$ Step 3. $$p+q=10 .$$ Check. $t_p+t_q=\tfrac52$ and $t_pt_q=1$ give $\{2,\tfrac12\}=\{t_4,t_6\}$, and $4+6=10$. Answer: 10