Let a, ar, ar^2, ldots be an infinite G.P. If | vedclass

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(D) The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r}$,where $|r| < 1$.Given $\sum_{n=0}^{\infty} ar^n = 57$,we have $\frac{a}{1-r} = 57$ $

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$31$

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(D) The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r}$,where $|r| Given $\sum_{n=0}^{\infty} ar^n = 57$,we have $\frac{a}{1-r} = 57$ $(I)$.Given $\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$,this is a $G.P.$ with first term $a^3$ and common ratio $r^3$.So,$\frac{a^3}{1-r^3} = 9747$ $(II)$.Dividing $(I)^3$ by $(II)$:$\frac{a^3}{(1-r)^3} 	imes \frac{1-r^3}{a^3} = \frac{57^3}{9747}$$\frac{1-r^3}{(1-r)^3} = \frac{185193}{9747} = 19$$\frac{(1-r)(1+r+r^2)}{(1-r)^3} = 19$$\frac{1+r+r^2}{(1-r)^2} = 19$$1+r+r^2 = 19(1-2r+r^2)$$1+r+r^2 = 19 - 38r + 19r^2$$18r^2 - 39r + 18 = 0$Dividing by $3$: $6r^2 - 13r + 6 = 0$$(2r-3)(3r-2) = 0$Since $|r| Substituting $r = \frac{2}{3}$ into $(I)$:$a = 57(1 - \frac{2}{3}) = 57 	imes \frac{1}{3} = 19$.Thus,$a + 18r = 19 + 18(\frac{2}{3}) = 19 + 12 = 31$.

Let $a, ar, ar^2, \ldots$ be an infinite $G.P.$ If $\sum_{n=0}^{\infty} ar^n = 57$ and $\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$,then $a + 18r$ is equal to:

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