if the sum of first 5 terms of a G.P. is 155, sum of last 5 terms is 39680,first term is 5 and last term is 20480. find the number of terms of the sequence.


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Question Number 29645 by gyugfeet last updated on 11/Feb/18

$i f t h e s u m o f f i r s t 5 t e r m s o f a G . P . i s 155, s u m o f l a s t 5 t e r m s i s 39680, f i r s t t e r m i s 5 a n d l a s t t e r m i s 20480 . f i n d t h e n u m b e r o f t e r m s o f t h e s e q u e n c e .$
	Answered by Rasheed.Sindhi last updated on 11/Feb/18

	$Let the common ratio of the GP is r$ $Sum of first five terms :$ $5 + 5 r + {5 r}^{2} + {5 r}^{3} + {5 r}^{4} = 155$ $Or 1 + r + r^{2} + r^{3} + r^{4} = 31 . . . . . . . . . . . . . . . A$ $Sum of last five terms (in reverse order) :$ $20480 + \frac{20480}{r} + \frac{20480}{r^{2}} + \frac{20480}{r^{3}} + \frac{20480}{r^{4}} = 39680$ $20480 (1 + \frac{1}{r} + \frac{1}{r^{2}} + \frac{1}{r^{3}} + \frac{1}{r^{4}}) = 39680$ $1 + \frac{1}{r} + \frac{1}{r^{2}} + \frac{1}{r^{3}} + \frac{1}{r^{4}} = \frac{39680}{20480}$ $\frac{1}{r^{4}} (r^{4} + r^{3} + r^{2} + r + 1) = \frac{39680}{20480}$ $But from A, r^{4} + r^{3} + r^{2} + r + 1 = 31$ $\frac{1}{r^{4}} (31) = \frac{39680}{20480}$ $r^{4} = \frac{20480 \times 31}{39680} = 16$ $r = \pm 2$ $Let the number of terms of GP is n$ $Last term = {a r}^{n - 1}$ $a = 5 (First term), r = \pm 2, Last term = 39680$ $5 {(\pm 2)}^{n - 1} = 20480$ ${(\pm 2)}^{n - 1} = 4096$ ${(\pm 2)}^{n - 1} = {(\pm 2)}^{12}$ $n - 1 = 12$ $n = 13$ $The GP contains 13 terms$
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