in a geometric series, the first term =a, common ratio=r. If S_n denotes the sum of the n terms and U_n =Σ_(n=1) ^n S_(n,) then rS_n +(1−r)U


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Question Number 38559 by nishant last updated on 27/Jun/18

$i n a g e o m e t r i c s e r i e s, t h e f i r s t t e r m$ $= a, c o m m o n r a t i o = r . I f S_{n} d e n o t e s$ $t h e s u m o f t h e n t e r m s a n d U_{n} = \underset{n = 1}{\sum^{n}} S_{n,}$ $t h e n {r S}_{n} + (1 - r) U_{n} e q u a l s t o$ $(a) 0 (b) n (c) n a (d) n a r$
	Answered by MrW3 last updated on 27/Jun/18

	$A_{n} = {a r}^{n - 1}$ $S_{n} = a + a r + . . . + {a r}^{n - 1}$ ${r S}_{n} = a r + {a r}^{2} + . . . + {a r}^{n - 1} + {a r}^{n}$ ${r S}_{n} = a + a r + {a r}^{2} + . . . + {a r}^{n - 1} + {a r}^{n} - a$ ${r S}_{n} = S_{n} + {a r}^{n} - a$ $\Rightarrow S_{n} = \frac{a (1 - r^{n})}{1 - r} = \frac{a}{1 - r} - \frac{a}{1 - r} \times r^{n}$ $U_{n} = \underset{k = 1}{\sum^{n}} S_{k} = \frac{a n}{1 - r} - \frac{a}{1 - r} (r + r^{2} + . . . + r^{n})$ $U_{n} = \frac{a n}{1 - r} - \frac{a}{1 - r} \times \frac{r (1 - r^{n})}{1 - r}$ $(1 - r) U_{n} = n a - \frac{a r (1 - r^{n})}{1 - r}$ ${r S}_{n} = \frac{a r (1 - r^{n})}{1 - r}$ $\Rightarrow {r S}_{n} + (1 - r) U_{n} = n a$ $\Rightarrow A n s w e r (c) i s r i g h t .$
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