Question

the sum of the first 5 and first 10 terms of a g.p. are respectively 16 and 3904. find common ratio​

Answer 1

Answer:  $\fontsize{18}{10}{\textup{\textbf{The required common ratio of the G.P. is 3.}}}$

Step-by-step explanation:

Let a and r represents the first term and common ratio of the given G.P.

Then, the sum of first n terms of the G.P. is given by

​$S_n=\dfrac{a(1-r^{n})}{1-r}.$

According to the given information, we have

$S_5=16\\\\\Rightarrow \dfrac{a(1-r^5)}{1-r}=16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$S_{10}=3904\\\\\Rightarrow \dfrac{a(1-r^{10})}{1-r}=3904~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Dividing equation (ii) by (i), we get

$\dfrac{1-r^{10}}{1-r^5}=\dfrac{3904}{16}\\\\\\\Rightarrow \dfrac{(1+r^5)(1+r^5)}{1-r^5}=244\\\\\Rightarrow 1+r^5=244\\\\\Rightarrow r^5=243\\\\\Rightarrow r^5=3^5\\\\\Rightarrow r=3.$

Thus, the required common ratio of the G.P. is 3.

Learn more #

Question :The sum of the first two terms of a g.p is 5/3 and the sum to infinity of the series is 3. the common ratio is ?

Link : https://brainly.in/question/3692108.