Question

Find the sum to n terms of 0.5 ++ 0.55 ++ 0.555 ………​

Answer 1

Answer:

here is ur answer

Step-by-step explanation:

1.605

Answer 2

$\purple{ \mathtt{ \underline{ \huge \fbox{ \underline{S} \: \underline{o} \: \underline{l} \: \underline{u} \: \underline{t} \: \underline{i} \: \underline{o} \: \underline{n} : \: </p><p>}}}}$

This is not a GP ., however , we can relate it to a GP by writing the terms as

$\sf \hookrightarrow S_{n} = 0.5 + 0.55 + 0.555 + ..... + n \: terms \\ \\ \sf \hookrightarrow S_{n} = 5(0.1 + 0.11 + 0.111 + ..... + n \: terms) \\ \\ \mathtt{ Multiplying \: numerator \: and \: denominator \: by \: 9 \: , \: we \: get}</p><p>\\ \\ \sf \hookrightarrow S_{n} = \frac{5}{9} (0.9 + 0.99 + 0.999 + ..... + n \: terms) \\ \\ \sf \hookrightarrow S_{n} = \frac{5}{9} \bigg \{(1 - 0.1)(1 - 0.01) + (1 - 0.001).....n \: terms) \bigg \} \\ \\ \sf \hookrightarrow S_{n} = \frac{5}{9} (1 + 1 + 1 + ...... + n \: terms) - (0.1 + 0.01 + 0.001 + ..... + n \: terms) \\ \\ \sf \hookrightarrow S_{n} = \frac{5}{9} \bigg(n - \frac{ \frac{1}{10}(1 - \frac{1}{ {10}^{n} } ) }{1 - \frac{1}{10} } \bigg) \\ \\ \sf \hookrightarrow S_{n} = \frac{5}{9} \bigg(n - \frac{ \frac{1}{10}(1 - \frac{1}{ {10}^{n} } ) }{ \frac{9}{10} } \bigg ) \\ \\ \sf \hookrightarrow S_{n} = \frac{5}{9} \bigg(n - \frac{1}{ \cancel{10}}(1 - \frac{1}{ {10}^{n} } ) \times \frac{\cancel{10} }{9} \bigg) \\ \\ \sf \hookrightarrow S_{n} = \frac{5}{9} \bigg(n - \frac{1}{9}(1 - \frac{1}{ {10}^{n} } ) \bigg)$

Hence , the sum of the sequence 0.5 , 0.55 , 0.555 ......... is

$\large \sf \fbox{S_{n} = \frac{5}{9} \bigg(n - \frac{1}{9}(1 - \frac{1}{ {10}^{n} } ) \bigg) }$