Question

the third term of G.P. is 3/4 and the sum of the first five terms is 32/33 times the sum of the first ten terms.Find the sum of the first four terms.​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{In a G.P}$

$\mathsf{t_3=\dfrac{3}{4}\;\&\;S_5=\dfrac{32}{33}\;S_{10}}$

$\underline{\textsf{To find:}}$

$\mathsf{S_4}$

$\underline{\textsf{Solution:}}$

$\underrline{\mathsf{Concept\;used:}}$

$\mathsf{The\;n\,th\;term\;of\;the\;G.P\;a,ar,ar^2.............\;is}$

$\boxed{\mathsf{t_n=ar^{n-1}}}$

$\mathsf{t_3=\dfrac{3}{4}}$

$\implies\mathsf{ar^2=\dfrac{3}{4}}$........(1)

$\mathsf{S_5=\dfrac{32}{33}\;S_{10}}$

$\implies\mathsf{\dfrac{a(r^5-1)}{r-1}=\dfrac{32}{33}{\times}\dfrac{a(r^{10}-1)}{r-1}}$

$\implies\mathsf{r^5-1=\dfrac{32}{33}{\times}((r^5)^2-1^2)}$

$\implies\mathsf{r^5-1=\dfrac{32}{33}{\times}(r^5-1)(r^5+1)}$

$\implies\mathsf{1=\dfrac{32}{33}{\times}(r^5+1)}$

$\implies\mathsf{\dfrac{33}{32}=r^5+1}$

$\implies\mathsf{\dfrac{33}{32}-1=r^5}$

$\implies\mathsf{\dfrac{1}{32}=r^5}$

$\implies\boxed{\mathsf{r=\dfrac{1}{2}}}$

$\mathsf{From\;(1)}$

$\mathsf{a\left(\dfrac{1}{4}\right)=\dfrac{3}{4}}$

$\implies\boxed{\mathsf{a=3}}$

$\textsf{Sum of first four terms}$

$\mathsf{S_n=\dfrac{a(r^n-1)}{r-1}}$

$\mathsf{S_4=\dfrac{a(r^4-1)}{r-1}}$

$\mathsf{S_4=\dfrac{3\left((\dfrac{1}{2})^4-1\right)}{\dfrac{1}{2}-1}}$

$\mathsf{S_4=\dfrac{3\left(\dfrac{1}{16}-1\right)}{\dfrac{-1}{2}}}$

$\mathsf{S_4=\dfrac{3\left(\dfrac{-15}{16}\right)}{\dfrac{-1}{2}}}$

$\mathsf{S_4=3\left(\dfrac{15}{8}\right)}$

$\therefore\boxed{\mathsf{S_4=\dfrac{45}{8}}}$

$\underrline{\textsf{Find more:}}$

The 3rd term of a GP is 2/3 and 6th term is 2/81 then the 1st term is

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