JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 1)
Let $${a_1}$$
, $${a_2}$$
, $${a_3}$$
,....... be a G.P. such that
$${a_1}$$ < 0, $${a_1}$$ + $${a_2}$$ = 4 and $${a_3}$$ + $${a_4}$$ = 16.
If $$\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $$, then $$\lambda $$ is equal to:
$${a_1}$$ < 0, $${a_1}$$ + $${a_2}$$ = 4 and $${a_3}$$ + $${a_4}$$ = 16.
If $$\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $$, then $$\lambda $$ is equal to:
171
-171
-513
$${{511} \over 3}$$
Explanation
$${a_1}$$
+ $${a_2}$$
= 4
$$ \Rightarrow $$ $${a_1}$$ + $${a_1}$$r = 4 ...(1)
$${a_3}$$ + $${a_4}$$ = 16
$$ \Rightarrow $$ $${a_1}$$r2 + $${a_1}$$r3 = 16 ...(2)
Doing (1) $$ \div $$ (2), we get
r = $$ \pm $$ 2
If r = 2, then a1 = $${4 \over 3}$$
If r = -2, then a1 = -4
Given $${a_1}$$ < 0
$$ \therefore $$ a1 = -4
$$ \therefore $$ $$\sum\limits_{i = 1}^9 {{a_i}}$$ = $${{a\left( {{r^9} - 1} \right)} \over {r - 1}}$$ = 4$$\lambda $$
$$ \Rightarrow $$ $${{ - 4\left( {{{\left( { - 2} \right)}^9} - 1} \right)} \over { - 2 - 1}}$$ = 4$$\lambda $$
$$ \Rightarrow $$ $$\lambda $$ = -171
$$ \Rightarrow $$ $${a_1}$$ + $${a_1}$$r = 4 ...(1)
$${a_3}$$ + $${a_4}$$ = 16
$$ \Rightarrow $$ $${a_1}$$r2 + $${a_1}$$r3 = 16 ...(2)
Doing (1) $$ \div $$ (2), we get
r = $$ \pm $$ 2
If r = 2, then a1 = $${4 \over 3}$$
If r = -2, then a1 = -4
Given $${a_1}$$ < 0
$$ \therefore $$ a1 = -4
$$ \therefore $$ $$\sum\limits_{i = 1}^9 {{a_i}}$$ = $${{a\left( {{r^9} - 1} \right)} \over {r - 1}}$$ = 4$$\lambda $$
$$ \Rightarrow $$ $${{ - 4\left( {{{\left( { - 2} \right)}^9} - 1} \right)} \over { - 2 - 1}}$$ = 4$$\lambda $$
$$ \Rightarrow $$ $$\lambda $$ = -171
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