JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 3)
If the sum of the second, third and fourth terms
of a positive term G.P. is 3 and the sum of its
sixth, seventh and eighth terms is 243, then the
sum of the first 50 terms of this G.P. is :
$${2 \over {13}}\left( {{3^{50}} - 1} \right)$$
$${1 \over {13}}\left( {{3^{50}} - 1} \right)$$
$${1 \over {26}}\left( {{3^{49}} - 1} \right)$$
$${1 \over {26}}\left( {{3^{50}} - 1} \right)$$
Explanation
Let first term = a > 0
Common ratio = r > 0
ar + ar2 + ar3 = 3 ....(i)
ar5 + ar6 + ar7 = 243 ....(ii)
$$ \Rightarrow $$ r4(ar + ar2 + ar3) = 243
$$ \Rightarrow $$ r4(3) = 243
$$ \Rightarrow $$ r = 3 as r > 0
from (i)
3a + 9a + 27a = 3
$$ \Rightarrow $$ a = $${1 \over {13}}$$
$$ \therefore $$ S50 = $${{a\left( {{r^{50}} - 1} \right)} \over {\left( {r - 1} \right)}}$$
= $${1 \over {26}}\left( {{3^{50}} - 1} \right)$$
Common ratio = r > 0
ar + ar2 + ar3 = 3 ....(i)
ar5 + ar6 + ar7 = 243 ....(ii)
$$ \Rightarrow $$ r4(ar + ar2 + ar3) = 243
$$ \Rightarrow $$ r4(3) = 243
$$ \Rightarrow $$ r = 3 as r > 0
from (i)
3a + 9a + 27a = 3
$$ \Rightarrow $$ a = $${1 \over {13}}$$
$$ \therefore $$ S50 = $${{a\left( {{r^{50}} - 1} \right)} \over {\left( {r - 1} \right)}}$$
= $${1 \over {26}}\left( {{3^{50}} - 1} \right)$$
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