Question

Find sum 2/5+3/5²+2/5³+3/5⁴+...........to infinity

Answer 1

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



$\sf = \dfrac{2}{5} \: + \: \dfrac{3}{5^2} \: + \: \dfrac{2}{5^3} \: + \: \dfrac{3}{5^4} \: + ..... \: \infty$


$\textsf{Rearranging the terms , } \\ \\ \sf = \: \left\{\dfrac{2}{5} \: + \: \dfrac{2}{5^3} \: + \: \dfrac{2}{5^5} \: + ..... \: \infty \right\} \: + \: \left\{\dfrac{3}{5^2} \: + \: \dfrac{3}{5^4} \: + \: \dfrac{3}{5^6} \: + ..... \: \infty \right\}$



$\sf = \: \left\{2 \underbrace{\left (\dfrac{1}{5} \: + \: \dfrac{1}{5^3} \: + \: \dfrac{1}{5^5} \: + ..... \: \infty \right) }_{This \: Part \: forms \: an \: infinite \: G.P.} \right\} \: + \: \left\{3 \underbrace{\left(\dfrac{1}{5^2} \: + \: \dfrac{1}{5^4} \: + \: \dfrac{1}{5^6} \: + ..... \: \infty \right)} _ {This \: Part \: too \: forms \: an \: infinite \: G.P.}\right\}$



$\underline{\textsf{For First G.P.,}} \\ \\ \sf \implies First \: term(a_1) \: = \: \dfrac{1}{5} \\ \\ \sf \implies Common \: ratio (r_1 ) \: = \: \dfrac{1}{5^3} \: \div \: \dfrac{1}{5} \\ \\ \sf \qquad \: \qquad \qquad \qquad \qquad \: = \dfrac{1}{5 {}^{3}} \: \times \: 5 \: = \: \dfrac{1}{ {5}^{2} }$



$\underline{\textsf{For Second G.P.,}} \\ \\ \sf \implies First \: term(a_2) \: = \: \dfrac{1}{ {5}^{2} } \\ \\ \sf \implies Common \: ratio (r_2 ) \: = \: \dfrac{1}{5^4} \: \div \: \dfrac{1}{ {5}^{2} } \\ \\ \sf \qquad \: \qquad \qquad \qquad \qquad \: = \dfrac{1}{5 {}^{4}} \: \times \: {5}^{2} \: = \: \dfrac{1}{ {5}^{2} }$




$\underline{\textsf{Using Formula : }} \\ \\ \sf \implies S_{\infty} \: = \: \dfrac{a}{1 \: - \: r } \quad \{ Where , \: r \: < \: |1| \}$




$\sf = \left\{ 2\left(\dfrac{\dfrac{1}{5}}{ 1 \: - \: \dfrac{1}{5^2}} \right) \right\} \: + \: \left\{ 3\left(\dfrac{\dfrac{1}{ {5}^{2} }}{ 1 \: - \: \dfrac{1}{5^2}} \right) \right\} \\ \\ \\ \sf = 2 \left( \dfrac{ \dfrac{1}{5} }{1 \: - \: \dfrac{1}{25} } \right) \: + \: 3 \left( \dfrac{ \dfrac{1}{25} }{1 \: - \: \dfrac{1}{25} } \right)$




$\sf = 2 \left( \dfrac{ \dfrac{1}{5} }{ \dfrac{25 \: - \: 1}{25} } \right) \: + \: 3 \left( \dfrac{ \dfrac{1}{25} }{\dfrac{25 \: - \: 1}{25} } \right) \\ \\ \\ \sf = 2 \left( \dfrac{ \quad\dfrac{1}{5} \quad}{ \dfrac{24}{25} } \right) \: + \: 3 \left( \dfrac{ \quad \dfrac{1}{ \cancel{25}} \quad }{\dfrac{24}{ \cancel{25} }} \right) \\ \\ \\ \sf = 2 \left( \dfrac{1}{5} \: \div \: \dfrac{24}{25} \right) \: + \: \dfrac{3}{24} \\ \\ \\ \sf = 2 \left( \dfrac{1}{5} \: \times \: \dfrac{25}{24} \right) \: + \: \dfrac{3}{24} \\ \\ \\ \sf = 2 \: \times \: \dfrac{5}{24} \: + \: \dfrac{3}{24}$




$\sf = \dfrac{10}{24} \: + \: \dfrac{3}{24} \\ \\ \sf = \dfrac{10 \: + \: 3 }{24} \\ \\ \sf = \dfrac{13}{24}$

Answer 2

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



$\sf = \dfrac{2}{5} \: + \: \dfrac{3}{5^2} \: + \: \dfrac{2}{5^3} \: + \: \dfrac{3}{5^4} \: + ..... \: \infty$


$\textsf{Rearranging the terms , } \\ \\ \sf = \: \left\{\dfrac{2}{5} \: + \: \dfrac{2}{5^3} \: + \: \dfrac{2}{5^5} \: + ..... \: \infty \right\} \: + \: \left\{\dfrac{3}{5^2} \: + \: \dfrac{3}{5^4} \: + \: \dfrac{3}{5^6} \: + ..... \: \infty \right\}$



$\sf = \: \left\{2 \underbrace{\left (\dfrac{1}{5} \: + \: \dfrac{1}{5^3} \: + \: \dfrac{1}{5^5} \: + ..... \: \infty \right) }_{This \: Part \: forms \: an \: infinite \: G.P.} \right\} \: + \: \left\{3 \underbrace{\left(\dfrac{1}{5^2} \: + \: \dfrac{1}{5^4} \: + \: \dfrac{1}{5^6} \: + ..... \: \infty \right)} _ {This \: Part \: too \: forms \: an \: infinite \: G.P.}\right\}$



$\underline{\textsf{For First G.P.,}} \\ \\ \sf \implies First \: term(a_1) \: = \: \dfrac{1}{5} \\ \\ \sf \implies Common \: ratio (r_1 ) \: = \: \dfrac{1}{5^3} \: \div \: \dfrac{1}{5} \\ \\ \sf \qquad \: \qquad \qquad \qquad \qquad \: = \dfrac{1}{5 {}^{3}} \: \times \: 5 \: = \: \dfrac{1}{ {5}^{2} }$



$\underline{\textsf{For Second G.P.,}} \\ \\ \sf \implies First \: term(a_2) \: = \: \dfrac{1}{ {5}^{2} } \\ \\ \sf \implies Common \: ratio (r_2 ) \: = \: \dfrac{1}{5^4} \: \div \: \dfrac{1}{ {5}^{2} } \\ \\ \sf \qquad \: \qquad \qquad \qquad \qquad \: = \dfrac{1}{5 {}^{4}} \: \times \: {5}^{2} \: = \: \dfrac{1}{ {5}^{2} }$




$\underline{\textsf{Using Formula : }} \\ \\ \sf \implies S_{\infty} \: = \: \dfrac{a}{1 \: - \: r } \quad \{ Where , \: r \: < \: |1| \}$




$\sf = \left\{ 2\left(\dfrac{\dfrac{1}{5}}{ 1 \: - \: \dfrac{1}{5^2}} \right) \right\} \: + \: \left\{ 3\left(\dfrac{\dfrac{1}{ {5}^{2} }}{ 1 \: - \: \dfrac{1}{5^2}} \right) \right\} \\ \\ \\ \sf = 2 \left( \dfrac{ \dfrac{1}{5} }{1 \: - \: \dfrac{1}{25} } \right) \: + \: 3 \left( \dfrac{ \dfrac{1}{25} }{1 \: - \: \dfrac{1}{25} } \right)$




$\sf = 2 \left( \dfrac{ \dfrac{1}{5} }{ \dfrac{25 \: - \: 1}{25} } \right) \: + \: 3 \left( \dfrac{ \dfrac{1}{25} }{\dfrac{25 \: - \: 1}{25} } \right) \\ \\ \\ \sf = 2 \left( \dfrac{ \quad\dfrac{1}{5} \quad}{ \dfrac{24}{25} } \right) \: + \: 3 \left( \dfrac{ \quad \dfrac{1}{ \cancel{25}} \quad }{\dfrac{24}{ \cancel{25} }} \right) \\ \\ \\ \sf = 2 \left( \dfrac{1}{5} \: \div \: \dfrac{24}{25} \right) \: + \: \dfrac{3}{24} \\ \\ \\ \sf = 2 \left( \dfrac{1}{5} \: \times \: \dfrac{25}{24} \right) \: + \: \dfrac{3}{24} \\ \\ \\ \sf = 2 \: \times \: \dfrac{5}{24} \: + \: \dfrac{3}{24}$




$\sf = \dfrac{10}{24} \: + \: \dfrac{3}{24} \\ \\ \sf = \dfrac{10 \: + \: 3 }{24} \\ \\ \sf = \dfrac{13}{24}$