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8. Sequences and Series
easy
अनुक्रम का कौन सा पद.
$2,2 \sqrt{2}, 4, \ldots ; 128$ है ?
A
$13^{\text {th }}$
B
$13^{\text {th }}$
C
$13^{\text {th }}$
D
$13^{\text {th }}$
Solution
The given sequence is $2,2 \sqrt{2}, 4 \ldots \ldots$ is $128 ?$
Here, $a=2$ and $r=(2 \sqrt{2}) / 2=\sqrt{2}$
Let the $n^{\text {th }}$ term of the given sequence be $128 .$
$a_{n}=a r^{n-1}$
$\Rightarrow(2)(\sqrt{2})^{n-1}=128$
$\Rightarrow(2)(2)^{\frac{n-1}{2}}=(2)^{7}$
$\Rightarrow(2)^{\frac{n-1}{2}+1}=(2)^{7}$
$\therefore \frac{n-1}{2}+1=7$
$\Rightarrow \frac{n-1}{2}=6$
$\Rightarrow n-1=12$
$\Rightarrow n=13$
Thus, the $13^{\text {th }}$ term of the given sequence is $128$
Standard 11
Mathematics
