The sum of the first five terms of the series | vedclass

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Question

(a) Given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + ........ = 3 + \frac{9}{2} + \frac{{27}}{4} + .....$

$ = 3 + \frac{{{3^2}}}{2} + \frac{{{3^3

Vedclass · Accepted Answer

$39\frac{9}{{16}}$

Vedclass · Answer

(a) Given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + ........ = 3 + \frac{9}{2} + \frac{{27}}{4} + .....$

$ = 3 + \frac{{{3^2}}}{2} + \frac{{{3^3}}}{4} + \frac{{{3^4}}}{8} + \frac{{{3^5}}}{{16}} + .....$(in $G.P.$)

Here $a = 3,\;r = \frac{3}{2}$ , then sum of the five terms

${S_5} = \frac{{a({r^n} - 1)}}{{r - 1}} = \frac{{3\left[ {{{\left( {\frac{3}{2}} ight)}^5} - 1} ight]}}{{\frac{3}{2} - 1}} = \frac{{1\left[ {\frac{{{3^5}}}{{32}} - 1} ight]}}{{\frac{1}{2}}}$

$ = 6\left[ {\frac{{243 - 32}}{{32}}} ight] = \frac{{211 	imes 3}}{{16}} = \frac{{633}}{{16}} = 39\frac{9}{{16}}$.

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + ......$ will be

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