If the ${10^{th}}$ term of a geometric progression is $9$ and ${4^{th}}$ term is $4$, then its ${7^{th}}$ term is
$A)\,6$
$B)\,36$
$C)\,\dfrac{4}{9}$
$D)\,\dfrac{9}{4}$

Answer
VerifiedVerified
159.6k+ views
Hint: First, we need to know about the concept of AM and GM. An arithmetic progression that can be given by $a,(a + d),(a + 2d),(a + 3d),...$ where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.
Hence the given question is in the form of geometric progression. And in the ${10^{th}}$ term of a geometric progression is $9$ and ${4^{th}}$ the term is $4$, we have to its ${7^{th}}$ term.
Formula used:
The general GP formula for the \[{n^{th}}\] term is given as ${a_n} = a{r^{n - 1}}$

Complete step-by-step solution:
Since from the given question we have ${10^{th}}$ term of a geometric progression is $9$. By use of the GP formula, we have written this into the equation of the ${10^{th}}$ term is ${a_{10}} = a{r^9} = 9$ where $n - 1 = 10 - 1 = 9$
Similarly, for the ${4^{th}}$ term is given as $4$ then we get ${a_4} = a{r^3} = 4$ where $n - 1 = 4 - 1 = 3$
Hence, we have two-equation, ${a_{10}} = a{r^9} = 9$ and ${a_4} = a{r^3} = 4$
Now let us multiply the two equations we get $(a{r^9} = 9) \times (a{r^3} = 4) \Rightarrow a{r^9}a{r^3} = 9 \times 4 = 36$
Since by the power rule concept we can write ${a^1} \times {a^1} = {a^{1 + 1}} = {a^2}$
Thus, we get ${a^2}{r^{12}} = 36$
Now taking the square terms common we have \[{a^2}{r^{12}} = 36 \Rightarrow {(a{r^6})^2} = 36\]
Further solving we get \[{(a{r^6})^2} = 36 \Rightarrow a{r^6} = \sqrt {36} = 6\]
Thus, we get \[a{r^6} = 6\]
Since the requirement is ${7^{th}}$ term and which can be generalized as ${a_7} = a{r^6}$ where $n - 1 = 7 - 1 = 6$
Hence, we get \[{a_7} = a{r^6} = 6\]
Therefore, the option $A)\,6$ is correct.

Note: Geometric Progression:
> In the GP the new series is obtained by multiplying the two consecutive terms so that they have constant factors.
> In GP the series is identified with the help of a common ratio between consecutive terms.
> Series vary in the exponential form because it increases by multiplying the terms.
For GP with the common ratio the formula to be calculated $GP = \dfrac{a}{{r - 1}},r \ne 1,r < 0$ and $GP = \dfrac{a}{{1 - r}},r \ne 1,r > 0$
Harmonic progress is the reciprocal of the given arithmetic progression which is the form of $HP = \dfrac{1}{{[a + (n - 1)d]}}$ where $a$ is the first term and $d$ is a common difference and n is the number of AP.