How many terms of the G.P. \( 3, \dfrac{3}{2}, \dfrac{3}{4}, …\) Be taken together to make \( \dfrac{3069}{512}\) ?

Asked by Sakshi | 1 year ago |  84

1 Answer

Solution :-

Given:

Sum of G.P = \(\dfrac{ 3069}{512}\)

Where, a = 3, r =\( \dfrac{ \dfrac{3}{2}}{3} = \dfrac{1}{2}\), n = ?

By using the formula,

Sum of GP for n terms = 

\( \dfrac{ 3069}{512}\)\(\dfrac{ 3( 9\dfrac{1}{2})n – 1)}{ (\dfrac{1}{2} – 1)}\)

\( \dfrac{ 3069}{512}×3×2 =\) \( 1- (\dfrac{1}{2})^n\)

\( \dfrac{ 3069}{3072}– 1=\)  \( – (\dfrac{1}{2})^n\)

\(\dfrac{ (3069 – 3072)}{3072} = – (\dfrac{1}{2})^n\)

10 = n

10 terms are required to make \( \dfrac{ 3069}{512}\)

Answered by Sakshi | 1 year ago

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