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in Geometric Progressions by (23.5k points)
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The sum of the first 20 terms of the sequence 0.7, 0.77, 0.777, ..... is

(a) \(\frac{7}{81}(179-10^{-20})\)

(b) \(\frac{7}{9}(99-10^{-20})\)

(c) \(\frac{7}{81}(179+10^{-20})\)

(d) \(\frac{7}{9}(99+10^{-20})\)

1 Answer

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Best answer

(c) \(\frac{7}{81}(179+10^{-20})\)

Let S20 = 0.7 + 0.77 + 0.777 + ..... upto 20 terms 

= 7(0.1 + 0.11 + 0.111 + ..... upto 20 terms) 

= \(\frac{7}{9}\) (0.9 + 0.99 + 0.999 + ..... upto 20 terms) 

= \(\frac{7}{9}\) [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) + upto 20 terms) 

= \(\frac{7}{9}\) [(1 + 1 + 1 + ..... upto 20 terms) – (0.1 + 0.01 + 0.001 + ..... upto 20 terms)]

\(\frac{7}{9}\)\(\bigg[20-\frac{0.1\{1-(0.1)^2\}}{(0-0.1)}\bigg]\)           \(\bigg(\because\,S_n=\frac{a(1-r^n)}{1-r}, \text{when}\,r<1\bigg)\)

\(\frac{7}{9}\)\(\bigg[20-\frac{1}{9}\bigg(1-\big({\frac{1}{10}\big)^{20}}\bigg)\bigg]\) = \(\frac{7}{9}\)\(\bigg[20-\frac{1}{9}+\frac{10}{9}^{-20}\bigg]\)

\(\frac{7}{9}\)\(\bigg[\frac{179+10^{-20}}{9}\bigg]\) = \(\frac{7}{81}(179+10^{-20})\).

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