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-

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 \dots \dots \dots \dots \dots 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