જો અનંત સમગુણોતર શ્રેણી GP : a, ar, ar^ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Sum of infinite terms :

$\frac{\mathrm{a}}{1-\mathrm{r}}=15....(i)$

Series formed by square of terms:

$\mathrm{a}^{2}, \mathrm{a}^{2} \mathrm

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

Sum of infinite terms :

$\frac{\mathrm{a}}{1-\mathrm{r}}=15....(i)$

Series formed by square of terms:

$\mathrm{a}^{2}, \mathrm{a}^{2} \mathrm{r}^{2}, \mathrm{a}^{2} \mathrm{r}^{4}, \mathrm{a}^{2} r^{6} \ldots \ldots$

$	ext { Sum }=\frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=150$

$\Rightarrow \frac{\mathrm{a}}{1-\mathrm{r}} \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150 \Rightarrow 15 \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150$

$\Rightarrow \frac{\mathrm{a}}{1+\mathrm{r}}=10 \ldots \ldots \ldots 	ext { (ii) }$

by $(i)$ and $(ii)$ $\mathrm{a}=12 ; \mathrm{r}=\frac{1}{5}$

Now series : $\mathrm{ar}^{2}, \mathrm{ar}^{4}, \mathrm{ar}^{6}$

$	ext { Sum }=\frac{a r^{2}}{1-r^{2}}=\frac{12 \cdot(1 / 25)}{1-1 / 25}=\frac{1}{2}$

Similar Questions

$2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots .$ to $\infty$ ની કિમંત મેળવો.

${{(0.2)}^{{{\log }_{\sqrt{5}}}\left( \frac{\text{1}}{\text{4}}\,+\,\frac{\text{1}}{\text{8}}\,+\,\frac{\text{1}}{\text{16}}\,+\,.....\,\infty \right)}}$ નું મૂલ્ય:

સમગુણોત્તર શ્રેણીમાં પ્રથમ $1$ છે. જો $4T_2 + 5T_3$ ન્યૂનત્તમ હોય, તો તેનો સામાન્ય ગુણોત્તર કેટલો થાય ?