Find the sum of first 20 terms of the geometric series 5/2+5/6+5/18+…..

Question

Find the sum of first 20 terms of the geometric series 5/2+5/6+5/18+…..

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Madelyn 3 weeks 2021-09-07T13:06:33+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-09-07T13:07:54+00:00

    r = t₂/t₁ a = 5/2 and n = 20

    r = (5/6)/(5/2)

    = (5/6) x (2/5)

    r = 1/3 here r < 1

    sn = a(1- rn)⁄(1 – r)

    So, S₂₀ = (5/2) [1- (1/3)^₂₀]/[1-(1/3)]

    = (5/2) [1- (1/3)^₂₀]/[2/3]

    = (5/2) x (3/2) [1- (1/3)^₂₀]

    = (15/4) [1 – (1/3)^₂₀]

    0
    2021-09-07T13:08:17+00:00

    GIVEN :

      \\ { \bold{ \dfrac{5}{2} +  \dfrac{5}{6} +  \dfrac{5}{18} + ....... }} \\

    TO FIND :

    Sum of first 20 terms = ?

    SOLUTION :

     \\ \:  \:  { \huge{.}} \:  \: { \bold{First \:  \: term \:  \: of \:  \: G.P.(a) =  \dfrac{5}{2} }} \\

     \\ \:  \:  { \huge{.}} \:  \: { \bold{Common \:  \: ratio \:  \: of \:  \: G.P.(r) =  \dfrac{1}{3} }} \\

     \\ \:  \:  { \huge{.}} \:  \: { \bold{n =  20}} \\

    • We know that –

     \\ \:  \:  { \huge{.}} \:  \: { \bold{Sum \:  \: of \:  \: G.P.=  \dfrac{a( {r}^{n} - 1) }{(r - 1)} }} \\

    • Now put the values –

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  \dfrac{ \left( \dfrac{5}{2} \right ) \left \{ {  \left(\dfrac{1}{3}  \right)}^{20} - 1  \right\} }{ \dfrac{1}{3}  - 1} }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  \dfrac{ \left( \dfrac{5}{2} \right ) \left \{ {  \dfrac{1}{{3}^{20}}}- 1  \right\} }{ -  \dfrac{2}{3}} }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  \dfrac{ \left( \dfrac{5}{2} \right ) \left \{ {1 -   \dfrac{1}{{3}^{20}}} \right\} }{   \dfrac{2}{3}} }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  \dfrac{ \left( \dfrac{5}{2} \right ) \left \{ {\dfrac{ {3}^{20}  - 1}{{3}^{20}}} \right\} }{   \dfrac{2}{3}} }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  { \left( \dfrac{5}{4} \right ) \left \{ {\dfrac{ {3}^{20}  - 1}{{3}^{19}}} \right\} } }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  { \left( \dfrac{5}{4} \right ) \left \{ {\dfrac{ 3,486,784,401 - 1}{1,162,261,467}} \right\} } }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  { \left( \dfrac{5}{4} \right ) \left \{ {\dfrac{ 3,486,784,400}{1,162,261,467}} \right\} } }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  {5 \left \{ {\dfrac{ 871,696,100}{1,162,261,467}} \right\} } }} \\

     \\ \implies { \bold{Sum \:  \: of \:  \: G.P.=  {\dfrac{ 4,358,480,500}{1,162,261,467 }}}} \\

     \\ \implies \large{ \boxed{ \bold{Sum \:  \: of \:  \: G.P.=3.75}}} \\

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