Question

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16. Sum of n term of the series 0.1 + 0.11 + 0.111 + ..... is -----


17. The sum of the first 20 terms of a G.P. is 245 times the sum of its first 10 terms . The common ratio is --


PLZ ANSWER IT .....

Answer 1

$\underline{\mathfrak{Solution : }}$


$\underline{\mathsf{Q.no \: 16 }}$


$\mathsf{ = 0.1 \: + \: 0.11 \: + \: 0.111 \: + \: ...... \: n \: terms} \\ \\ \mathsf{ = 0.1(1 \: + \: 1.1 \: + \: 1.11 \: + \: ...... \: n \: terms)}$

$\mathsf{Multiply \: the \: numerator\: and \: denominator \: by \: 9, }$


$\mathsf{ = \dfrac{1}{10 \times 9}(1 \: + \: 1.1 \: + \: 1.11 \: ...... \: n \: terms )9} \\ \\ \\ \mathsf{ = \dfrac{1}{90}(9 \: + \: 9.9 \: + \: 9.99 ...... \: n \: terms)} \\ \\ \\ \mathsf{ = \dfrac{1}{90} [(10 \: - \:1) \: + \: ( 10 \: - \: 0.1) \: + \: ( 10 \: - \: 0.01) ...... \: n \: terms]}$


$\\ \mathsf{ = \dfrac{1}{90}[10n \: - \:( 1 \: + \: 0.1 \: + \: 0.01...... \: n \: terms)]}$



$\mathsf{Now, \: for \: the \: G.P. \: part, } \\ \\ \mathsf{\implies First \: term (a)\: = \: 1 } \\ \\ \mathsf{\implies Common \: ratio (r) \: = \: \dfrac{0.1}{1} \: = \: 0.1 \: }$




$\mathsf{Since, \: r \: < \: 1} \\ \\ \mathsf{So, \: sum \: of \: n \: terms \: = \: \dfrac{a(1 \: - \: {r}^{n})}{(1 \: - \: r )}} \\ \\ \\ \mathsf{ \qquad \qquad \qquad \qquad \: = \: \dfrac{1[1 \: - \: {(0.1)}^{n}] }{(1 \: - \: 0.1)} } \\ \\ \\ \mathsf{\qquad \qquad \qquad \qquad \: = \: \dfrac{(1 \: - \: {(0.1 )}^{n} ) }{0.9}}$



$\underline{\mathsf{Now,}}$



$\mathsf{ = \: \dfrac{1}{90}[10n \: - \: \dfrac{1 \: - \: ({0.1})^{n}}{0.9}]} \\ \\ \\ \mathsf{ = \: \dfrac{1}{90} \: \times \: \dfrac{[10n \: \times \: 0.9 \: - ( 1 \: - \: (0.1)^{n})]}{0.9}}$


$\\ \mathsf{= \: \dfrac{1}{90} \: \times \: \dfrac{[10n \: \times \: 0.9 \: - \:( 1 \: - \: (0.1)^{n}] }{\dfrac{9}{10}}}$



$\\ \mathsf{= \: \dfrac{1 0}{90} \: \times \: [\dfrac{ \cancel{10}n\: \times \: \dfrac{9}{ \cancel{10}} \: - \: ( 1 \: - \: (0.1)^{n})}{9}]}$




$\\ \mathsf{= \: \dfrac{1}{9} \: \times \: [\dfrac{9 \: - \: ( 1 \: - \: (0.1)^{n})}{9}]}$




$\mathsf{= \: \dfrac{1}{9}\: \times \: [\dfrac{9}{9} \: - \:\dfrac{(1 \: - \: (0.1)^{n})}{9}]}$



$\mathsf{= \: \dfrac{1}{9}[1 \: - \:(1 \: - \:(0.1)^{n})/9]}$



$\boxed{\underline{\mathsf{The \: required \: answer \: is \: (b).}}}$




$\underline{\mathsf{Q. no \: 17 }}$



$\mathsf{We \: know \: that,}$


$\\ \mathsf{\implies Sum \: of \: n \: terms \: = \: \dfrac{a(r^{n} \: - \: 1)}{(r\: - \: 1)}}$


$\mathsf{Let,} \\ \\ \mathsf{\implies First \: term \: = \:a } \\ \\ \mathsf{\implies Common \: difference \: = \: d }$



$\mathsf{According \: to \: the \: question, } \\ \\ \\ \mathsf{\implies S_{20} \: = \: 244 \: \times \: S_{10}} \\ \\ \\ \mathsf{ \implies \dfrac{ \cancel{a}( r^{20 } \: - \: 1 )}{ \cancel{(r \: - \: 1)}} \: = \: 244 \: \times \: \dfrac{ \cancel{a}( {r}^{10} \: - \: 1)}{ \cancel{(r \: - \: 1)}}}$



$\\ \mathsf{\implies (r^{20} \: - \: 1) \: = \: 244 \: \times \: (r^{10} \: - \: 1)} \\ \\ \\ \mathsf{\implies \dfrac{(r^{20} \: - \:1)}{(r^{10} \: - \: 1)} \: = \: 244}$



$\\ \mathsf{\implies \dfrac{[(r^{10})^{2} \: - \: ({1})^{2}]}{(r^{10} \: - \: 1)} \: = \: 244 }$





$\\ \mathsf{Using \: Algebraic \: identity,} \\ \\ \\ \boxed{\mathsf{\implies (a)^{2} \: - \: (b)^{2} \: = \: (a \: + \: b )(a \: - \: b}}$



$\\ \mathsf{\implies \dfrac{(r^{10} \: + \: 1)\cancel{(r^{10} \: - \: 1) }}{\cancel{(r^{10} \: - \: 1)} }\: = \: 244 } \\ \\ \\ \mathsf{ \implies {r}^{10} \: + \: 1 \: = \: 244}$




$\\ \mathsf{\implies r^{10} \: = \: 244 \: - \: 1} \\ \\ \mathsf{\implies r^{10} \: = \: 243} \\ \\ \mathsf{\implies r^{10} \: = \: 3^{5}} \\ \\ \mathsf{ \implies {r}^{ \normalsize{\frac{10}{5}} } \: = \: 3} \\ \\ \mathsf{ \implies {r}^{2} \: = \: 3} \\ \\ \mathsf{ \therefore \quad r \: = \: \pm \sqrt{3}}$



$\boxed{\underline{\mathsf{The \: required \: answer \: is \: (a).}}}$

Answer 2

Step-by-step explanation:

.. Ca Mohammad Hamza Qureshi

Mumbai