Question

Third term of a GP is the square of its first term. If its second term is 8, then its 6th (a) 120 (b) 124 (c) 128 (d) 132​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ 6th \: term \: of \: GP \\ \\ so \: let \: third \: term \: of \: GP \: be \: ar {}^{2} \\ and \: its \: second \: term \: be \: ar \\ \\ ar {}^{2} = a {}^{2} \\ r { }^{2} = a \: \: \: \: \: \: \: (1) \\ \\ ar = 8 \\ r {}^{2} .r = 8 \\ r {}^{3} = 8 \\ r = 2 \\ \\ thus \: then \\ a = 4$

$so \: we \: know \: that \\ nth \: term \: of \: GP \: is \: given \: by \\ \\ tn = ar {}^{n - 1} \\ for \: \: n = 6 \\ \\ = 4.(2) {}^{6 - 1} \\ = 4.(2) {}^{5} \\ = 4 \times 32 \\ ar {}^{5} = 6th \: term = 128$