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

# Find the sum of the following geometric progressions: (i) 2, 6, 18, ... to 7 terms; (ii) 1, 3, 9, 27, ... to 8 terms; (iii) 1, −1/2, 1/4, −1/8, ... to 9 terms; (iv) (a2 − b2), (a − b), $\left(\frac{a-b}{a+b}\right)$, ... to n terms; (v) 4, 2, 1, 1/2 ... to 10 terms.

Open in App
Solution

## (i) Here, a = 2 and r = 3. $\therefore {S}_{7}=a\left(\frac{{r}^{7}-1}{r-1}\right)\phantom{\rule{0ex}{0ex}}=2\left(\frac{{3}^{7}-1}{3-1}\right)\phantom{\rule{0ex}{0ex}}=2187-1\phantom{\rule{0ex}{0ex}}=2186$ (ii) Here, a = 1 and r = 3. $\therefore {S}_{8}=a\left(\frac{{r}^{8}-1}{r-1}\right)\phantom{\rule{0ex}{0ex}}=1\left(\frac{{3}^{8}-1}{3-1}\right)\phantom{\rule{0ex}{0ex}}=\frac{6561-1}{2}\phantom{\rule{0ex}{0ex}}=3280$ (iii) Here, a = 1 and r = −$\frac{1}{2}$. $\therefore {S}_{9}=a\left(\frac{1-{r}^{9}}{1-r}\right)\phantom{\rule{0ex}{0ex}}=1\left(\frac{1-{\left(-\frac{1}{2}\right)}^{9}}{1-\left(-\frac{1}{2}\right)}\right)\phantom{\rule{0ex}{0ex}}=\frac{1-\left(-\frac{1}{512}\right)}{\frac{3}{2}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{513}{512}}{\frac{3}{2}}\phantom{\rule{0ex}{0ex}}=\frac{513×2}{512×3}\phantom{\rule{0ex}{0ex}}=\frac{171}{256}$ (iv) Here, a = a2 − b2 and r = $\frac{1}{a+b}$. $\therefore {S}_{n}=a\left(\frac{1-{r}^{n}}{1-r}\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{2}-{b}^{2}\right)\left(\frac{1-{\left(\frac{1}{a+b}\right)}^{n}}{1-\left(\frac{1}{a+b}\right)}\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{2}-{b}^{2}\right)\left(\frac{\left(\frac{{\left(a+b\right)}^{n}-1}{{\left(a+b\right)}^{n}}\right)}{\frac{\left(a+b\right)-1}{a+b}}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{\left(a+b\right)\left(a-b\right)}{{\left(a+b\right)}^{n-1}}\left(\frac{{\left(a+b\right)}^{n}-1}{\left(a+b\right)-1}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(a-b\right)}{{\left(a+b\right)}^{n-2}}\left(\frac{{\left(a+b\right)}^{n}-1}{\left(a+b\right)-1}\right)$ (v) Here, a = 4 and r = $\frac{1}{2}$ $\therefore {S}_{n}=a\left(\frac{1-{r}^{10}}{1-r}\right)\phantom{\rule{0ex}{0ex}}=4\left(\frac{1-{\left(\frac{1}{2}\right)}^{10}}{1-\left(\frac{1}{2}\right)}\right)\phantom{\rule{0ex}{0ex}}=4\left(\frac{1-\left(\frac{1}{1024}\right)}{\frac{1}{2}}\right)\phantom{\rule{0ex}{0ex}}=8\left(1-\frac{1}{1024}\right)\phantom{\rule{0ex}{0ex}}=\frac{1023}{128}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Formula for Sum of N Terms of an AP
MATHEMATICS
Watch in App
Join BYJU'S Learning Program