Question

If the \[{5^{th}}\] term of a G.P. is \[\frac{1}{3}\] and \[{9^{th}}\] term is \[\frac{{16}}{{243}}\], then the \[{4^{th}}\] term will be
A. \[\frac{3}{4}\]
B. \[\frac{1}{2}\]
C. \[\frac{1}{3}\]
D. \[\frac{2}{5}\]

Answer 1

Hint:
Given the fifth term and the ninth of the G.P., we have to find the product of fourth term. If we take the first term as ‘a’ and the common ratio of the entire G.P. as r, we can simply equate the required terms. The term to be found will then be multiplied so that their result is \[a,ar,a{r^3},a{r^3},......a{r^n}\]. So let's get to work
Formula use:
 first term as ‘a’ and the common ratio of the entire G.P. as r,\[{n^{th}}\] of G.P
\[{T_n} = a{r^{n - 1}}\]
Complete step-by-step solution
We have been given in the problem that,
Fifth term of a G.P. is\[\frac{1}{3}\]
Ninth term of a G.P is\[\frac{{16}}{{243}}\]
Now, let us have first term as ‘a’ and the common ratio of the whole G.P as ‘r’
Let’s use the formula,
\[{T_n} = a{r^{n - 1}}\]
For the fifth term, let’s substitute the value from the given question, we get
\[{T_5} = a{r^{5 - 1}} = \frac{1}{3}\]
Now, simplify the power of \[ar\] from the above equation,
\[{T_5} = a{r^4} = \frac{1}{3}\]------ (1)
For the ninth term, let’s substitute the value from the given question, we get
\[{T_9} = a{r^{9 - 1}} = \frac{{16}}{{243}}\]
Now, simplify the power of \[ar\] from the above equation,
\[{T_9} = a{r^8} = \frac{{16}}{{243}}\]---- (2)
Now, let us solve the equation (1) and equation (2), we obtain
\[{r^4} = \frac{{(a{r^8})}}{{(a{r^4})}}\]
\[ = \left( {\frac{{16}}{{243}}} \right) \times \left( {\frac{3}{1}} \right)\]
\[ = \left( {\frac{{16}}{{81}}} \right) = {\left( {\frac{2}{3}} \right)^4}\]
And,
\[a = \frac{{\left( {\frac{1}{3}} \right)}}{{{r^4}}} = \frac{{\frac{1}{3}}}{{{{\left( {\frac{2}{3}} \right)}^4}}} = \frac{{{3^3}}}{{{2^4}}}\]
Finally, fourth term is,
\[a{r^3} = \frac{{{3^3}}}{{{2^4}}} \cdot \frac{{{2^3}}}{{{3^3}}}\]
\[ = \frac{1}{2}\]
Therefore, the \[{4^{th}}\] term will be \[\frac{1}{2}\]
Hence, the option B is correct.
Note:
Student should be careful in finding terms, because dealing problems with powers are done with caution, because it gives wrong solution. Take note that the fifth term has a fourth power of 2 rather than a fifth power. A common ratio exists between terms in a G.P. That is why the power of r increases as the words increase.