Calculate the sum of the infinite geometric series 2-1-2-1-8-1-32-

The correct option is D 13/5 Given the geometric series is #160; 2, ( 12 ), ( 18 ), ( 132 )....................... Then common ratio = 14 And ...

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Standard X

Mathematics

Sum of Infinite Terms of a GP

Calculate the...

Question

Calculate the sum of the infinite geometric series $2 + (- \frac{1}{2}) + (\frac{1}{8}) + (- \frac{1}{32}) + . . .$

1 \frac{3}{8}

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1 \frac{2}{5}

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1 \frac{1}{2}

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1 \frac{3}{5}

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1 \frac{5}{8}

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Solution

The correct option is D $1 \frac{3}{5}$
Given the geometric series is $2, (- \frac{1}{2}), (\frac{1}{8}), (- \frac{1}{32}) . . . . . . . . . . . . . . . . . . . . . . .$
Then common ratio $= - \frac{1}{4}$
And first term is $2$ .
Then sum of the infinite geometric series $=$ $S = \frac{a_{1}}{1 - r} = \frac{2}{1 - (- \frac{1}{4})} = \frac{2 \times 4}{4 + 1} = \frac{8}{5} = 1 \frac{3}{8}$

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Similar questions

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

Q. Find the sum of the infinite geometric series

1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . . . . .

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

Q. Simplify:
(i)

\frac{11}{12} \div (\frac{5}{9} \times \frac{18}{25})

(ii)

(2 \frac{1}{2} \times \frac{8}{10}) \div (1 \frac{1}{2} + \frac{5}{8})

(iii)

\frac{15}{16}

(\frac{5}{6} - \frac{1}{2}) \div \frac{10}{11}

(iv)

\frac{9}{8} \div \frac{3}{5}

(\frac{3}{4} + \frac{3}{5})

(v)

\frac{2}{5} \div {\frac{1}{5} o f [\frac{3}{4} - \frac{1}{2}] - 1}

(vi)

(1 \frac{3}{4} \times 3 \frac{1}{7}) - (4 \frac{3}{8} \div 5 \frac{3}{5})

In this case how many time intervals can Sid calculate?

$\begin{matrix} B u s & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 T i m e (h r : m i n) & 8 : 15 & 8 : 18 & 8 : 23 & 8 : 29 & 8 : 32 & 8 : 40 & 8 : 45 & 8 : 52 & 8 : 54 & 9 : 00 & 9 : 07 & 9 : 13 & 9 : 20 & 9 : 25 & 9 : 32 \end{matrix}$

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Calculate the sum of the infinite geometric series 2+(−12)+(18)+(−132)+...

The correct option is D 135Given the geometric series is 2,(−12),(18),(−132).......................Then common ratio =−14And first term is 2.Then sum of the infinite geometric series = S=a11−r=21−(−14)=2×44+1=85=138

Calculate the sum of the infinite geometric series $2 + (- \frac{1}{2}) + (\frac{1}{8}) + (- \frac{1}{32}) + . . .$