Question

If the 4th and 7th terms of a G.P are 54 and 1458 respectively find the G.P

Answer 1

hey!!
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Here is your answer!!
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ar^3=54
ar^6=1458
ar^6/ar^3=1458/54
r^3=27
r=3
G.P. =a,ar,ar^2
ar^3=54
a3^3=54
a×27=54
a=54/27
a=2
G.P.=2,6,18.......
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I hope it will help you!!

Answer 2

Step-by-step explanation:

$Helloo.... \\ Tn = a {r}^{n - 1} \\ \\ T4 = 54 \\ a {r}^{3} = 54...........(1) \\ T7 = 1458 \\ a {r}^{6} = 1458.........(2) \\ \\ divide \: (2) \: by \: (1) = > \\ \frac{a {r}^{6} }{a {r}^{3} } = \frac{1458}{54} \\ {r}^{3} = 27 \\ {r}^{ 3} = {3}^{3} \\ r = 3 \\ put \: r = 3 \: in \: (1) = > \\ a {r}^{3} = 54 \\ 27a = 54 \\ a = 2 \\ \\ so \: your \: G. P\: will \: be = > \\ a, \: ar ,\: a {r}^{2}, ....... \\ 2 ,\: 6 ,\: 18........$