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Q.
How many digits will be there after decimal point in $ 12^{th} $ term of the sequence : $ 32 $ , $ 16 $ , $ 8 $ , $ 4 $ ,...... ?
J & K CETJ & K CET 2018
Solution:
We have $32, 16, 8, 4 ....$ are in $G.P.$
$a = 32, r = \frac{1}{2}$
Then, $12^{th}$ term of the sequence $ =ar^{12-1}$
$ = 32 \times \frac{1}{2^{11}} = 2^5 \times 2^{-11} = 2^{-6}$
$ = 0.015625$
Hence, six digits will be there after decimal point in $12^{th}$ term of the sequence.