JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 19)
If the arithmetic mean and geometric mean of the pth and qth terms of the
sequence $$-$$16, 8, $$-$$4, 2, ...... satisfy the equation
4x2 $$-$$ 9x + 5 = 0, then p + q is equal to __________.
sequence $$-$$16, 8, $$-$$4, 2, ...... satisfy the equation
4x2 $$-$$ 9x + 5 = 0, then p + q is equal to __________.
Answer
10
Explanation
Given, $$4{x^2} - 9x + 5 = 0$$
$$ \Rightarrow (x - 1)(4x - 5) = 0$$
$$ \Rightarrow $$ A. M. $$ = {5 \over 4}$$, G. M. = 1 (As A. M. $$ \ge $$ G. M)
Again, for the series
$$-$$16, 8, $$-$$4, 2 ..........
$${p^{th}}$$ term $${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$$
$${q^{th}}$$ term $${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$$
Now, A. M. = $${{{t_p} + {t_q}} \over 2} = {5 \over 4}$$ & G. M. = $$\sqrt {{t_p}{t_q}} = 1$$
$$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$$
$$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$$
$$ \Rightarrow p + q = 10$$
$$ \Rightarrow (x - 1)(4x - 5) = 0$$
$$ \Rightarrow $$ A. M. $$ = {5 \over 4}$$, G. M. = 1 (As A. M. $$ \ge $$ G. M)
Again, for the series
$$-$$16, 8, $$-$$4, 2 ..........
$${p^{th}}$$ term $${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$$
$${q^{th}}$$ term $${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$$
Now, A. M. = $${{{t_p} + {t_q}} \over 2} = {5 \over 4}$$ & G. M. = $$\sqrt {{t_p}{t_q}} = 1$$
$$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$$
$$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$$
$$ \Rightarrow p + q = 10$$
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