1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# How many terms of the G.P. 3, $\frac{3}{2},\frac{3}{4}$..... are needed to give the sum $\frac{3069}{512}$?

Open in App
Solution

## $\mathrm{Here},a=3\mathrm{and}\phantom{\rule{0ex}{0ex}}\mathrm{Common}\mathrm{ratio},r=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{And},{S}_{n}=\frac{3069}{512}\phantom{\rule{0ex}{0ex}}\therefore {S}_{n}=3\left\{\frac{1-{\left(\frac{1}{2}\right)}^{n}}{1-\frac{1}{2}}\right\}\phantom{\rule{0ex}{0ex}}⇒\frac{3069}{512}=3\left\{\frac{1-\frac{1}{{2}^{n}}}{1-\frac{1}{2}}\right\}\phantom{\rule{0ex}{0ex}}⇒\frac{3069}{512}=6\left\{1-\frac{1}{{2}^{n}}\right\}\phantom{\rule{0ex}{0ex}}⇒\frac{3069}{3072}=1-\frac{1}{{2}^{n}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{2}^{n}}=1-\frac{3069}{3072}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{2}^{n}}=\frac{3}{3072}\phantom{\rule{0ex}{0ex}}⇒{2}^{n}=\frac{3072}{3}\phantom{\rule{0ex}{0ex}}⇒{2}^{n}=1024\phantom{\rule{0ex}{0ex}}⇒{2}^{n}={2}^{10}\phantom{\rule{0ex}{0ex}}\therefore n=10$

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Sum of n Terms
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program