in-a-geometric-series-the-first-term-a-common-ratio-r-If-S-n-denotes-the-sum-of-the-n-terms-and-U-n-n-1-n-S-n-then-rS-n-1-r-U-n-equals-to-a-0-b-n-c-na-d-na

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 + \dots + {a r}^{n - 1}

{r S}_{n} = a r + {a r}^{2} + \dots + {a r}^{n - 1} + {a r}^{n}

{r S}_{n} = a + a r + {a r}^{2} + \dots + {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} + \dots + 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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