the sum to infinity of a Geometric series is S the sum to infinty of the squares of the terms of the series is 2S the sum to infinity of the cubes of the terms of the series is ((64)/(13))S. find the value of S and write iut the first 3 terms if the series.


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 78493 by Rio Michael last updated on 18/Jan/20

$t h e s u m t o i n f i n i t y o f a G e o m e t r i c s e r i e s i s S$ $t h e s u m t o i n f i n t y o f t h e s q u a r e s o f t h e t e r m s$ $o f t h e s e r i e s i s 2 S$ $t h e s u m t o i n f i n i t y o f t h e c u b e s o f t h e t e r m s$ $o f t h e s e r i e s i s \frac{64}{13} S .$ $f i n d t h e v a l u e o f S a n d w r i t e i u t t h e f i r s t$ $3 t e r m s i f t h e s e r i e s .$
	Answered by mr W last updated on 18/Jan/20

	$S = \frac{a_{1}}{1 - q} . . . (i)$ $2 S = \frac{a_{1}^{2}}{1 - q^{2}} > 0 . . . (i i)$ $\frac{64}{13} S = \frac{a_{1}^{3}}{1 - q^{3}} . . . (i i i)$ $a_{1} = (1 - q) S$ $2 S = \frac{a_{1}^{2}}{1 - q^{2}} = \frac{{(1 - q)}^{2} S^{2}}{1 - q^{2}} = \frac{1 - q}{1 + q} S^{2}$ $2 = \frac{1 - q}{1 + q} S$ $\Rightarrow q = \frac{S - 2}{S + 2} = 1 - \frac{4}{S + 2}$ $\Rightarrow a_{1} = \frac{4 S}{S + 2}$ $\frac{64}{13} S = \frac{a_{1}^{3}}{1 - q^{3}} = \frac{{(\frac{4 S}{S + 2})}^{3}}{1 - {(\frac{S - 2}{S + 2})}^{3}}$ $\frac{64}{13} S = \frac{4^{3} S^{3}}{{(S + 2)}^{3} - {(S - 2)}^{3}} = \frac{64 S^{3}}{2 (6 S^{2} + 8)}$ $\frac{1}{13} = \frac{S^{2}}{2 (6 S^{2} + 8)}$ $S^{2} = 16$ $\Rightarrow S = 4$ $\Rightarrow q = \frac{S - 2}{S + 2} = \frac{4 - 2}{4 + 2} = \frac{1}{3}$ $\Rightarrow a_{1} = \frac{4 S}{S + 2} = \frac{16}{6} = \frac{8}{3}$ $\Rightarrow a_{2} = \frac{8}{3} \times \frac{1}{3} = \frac{8}{9}$ $\Rightarrow a_{3} = \frac{8}{9} \times \frac{1}{3} = \frac{8}{27}$
	Commented by peter frank last updated on 18/Jan/20

	$t h a n k y o u$
	Answered by john santu last updated on 18/Jan/20

	$a + a r + {a r}^{2} + {a r}^{3} + . . . = \frac{a}{1 - r} = S \Rightarrow a = S (1 - r)$ $a^{2} + a^{2} r^{2} + a^{2} r^{4} + a^{2} r^{6} + . . . = \frac{a^{2}}{1 - r^{2}} = 2 S$ $\frac{a}{1 + r} \times \frac{a}{1 - r} = 2 S$ $\frac{a}{1 + r} = 2 \Rightarrow a = 2 + 2 r \Rightarrow S - S r = 2 + 2 r$ $a^{3} (1 + r^{3} + r^{6} + r^{9} + . . .) = \frac{64}{13} S$ $\frac{a^{3}}{1 - r^{3}} = \frac{64}{13} S \Rightarrow \frac{a}{(1 - r)} \times \frac{a^{2}}{(1 + r + r^{2})} = \frac{64}{13} S$ $13 a^{2} = 64 (1 + r + r^{2})$ $13 \times 4 (r^{2} + 2 r + 1) = 64 (r^{2} + r + 1)$ $13 r^{2} + 26 r + 13 = 16 r^{2} + 16 r + 16$ $3 r^{2} - 10 r + 3 = 0 \Rightarrow (3 r - 1) (r - 3) = 0$ $r = 3 (n o s o l u t i o n)$ $r = \frac{1}{3} \Rightarrow a = 2 + \frac{2}{3} = \frac{8}{3}$ $S = \frac{(8 / 3)}{1 - (1 / 3)} = 4$
	Commented by peter frank last updated on 18/Jan/20

	$t h a n k y o u$
	Commented by peter frank last updated on 18/Jan/20

	$p l e a s e h e l p Q n 77990$
	Commented by Rio Michael last updated on 18/Jan/20

	$t h a n k y o u s i r s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum