Question

answer is 5, with full explanation​

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\which \: \: term \: of \: given \: sequence \\ is \: \: \frac{1}{32} \: \: (ie \: \: n)\\ \\ so \: given \: sequence \: is \\ \\ 2 + \frac{1}{ \sqrt{2} } + \frac{1}{4} + so \: \: on \\ \\ so \: then \\ t1 = 2 \: \: \\ \\ t2 = \frac{1}{ \sqrt{2} } \\ \\ t 3 = \frac{1}{4} \\ \\ \frac{t2}{t1} = \frac{ \frac{1}{ \sqrt{2} } }{2} \\ \\ \frac{t2}{t1} = \frac{1}{2 \sqrt{2} } \\ \\ \frac{t3}{t2} = \frac{ \frac{ \frac{1}{4 } }{1} }{ \sqrt{2} } \\ \\ = \frac{ \sqrt{2} }{4} \\ \\ = \frac{ \sqrt{2} }{4} \times \frac{ \sqrt{2} }{ \sqrt{2} } \\ \\ = \frac{2}{4 \sqrt{2} } \\ \\ \frac{t3}{t2} = \frac{1}{2 \sqrt{2} }$

$so \: here \\ since \: common \: ratio \: (r) \: is \: constant \\ given \: sequence \: is \: G.P \\ \\ where \\ a = 2 \\ r = \frac{1}{2 \sqrt{2} }$

$so \: let \: then \\ tn = \frac{1}{32} \\ \\ we \: have \\ \\ tn = a.r {}^{n - 1} \\ \\ \frac{1}{32} = 2. (\frac{1}{2 \sqrt{2} } \: ) {}^{n - 1} \\ \\ ( \frac{1}{2 \sqrt{2} } \: ) {}^{n - 1} = \frac{1}{64} \\ \\ ( \frac{1}{2 \sqrt{2} } \: ) {}^{n - 1} = ( \frac{1}{2 \sqrt{2} } ) {}^{4} \\ \\ comparing \: respective \: exponentials \\ and \: bases \\ \\ we \: get \\ n - 1 = 4 \\ \\ n = 5$

$hence \: for \: given \: \: G.P \\ \frac{1}{32} \: \: is \: our \: \: 5th \: term$