Question

For a geometric sequence, ${{t}_{8}}=60\And {{t}_{6}}=72$, how do you find ${{t}_{10}}$?

Answer 1

We know that the general term in the geometric sequence is as follows: ${{t}_{n}}=a{{r}^{n-1}}$. In this equation, “a” is the first term of the sequence and “r” is the common ratio of the sequence. Then substitute the value of n as 8 and the general term corresponding to n equals 8 in the given equation. Then substitute n as 6 and ${{t}_{n}}=72$ in the given equation and then solve the two equations to get the value of “a & r” then we will find the value of ${{t}_{10}}$.

Complete step-by-step solution:
The general term in the geometric sequence is as follows:
${{t}_{n}}=a{{r}^{n-1}}$ ………… (1)
In the above formula for the general term, “a” is the first term of the sequence and “r” is the common ratio of the sequence.
Now, substituting n as 8 in the above general term and ${{t}_{8}}=60$ we get,
$\begin{aligned} & \Rightarrow {{t}_{8}}=a{{r}^{8-1}}=60 \\\ & \Rightarrow {{t}_{8}}=a{{r}^{7}}=60......(2) \\\ \end{aligned}$
Substituting n as 6 in eq. (1) and ${{t}_{6}}=72$ in eq. (1) we get,
$\begin{aligned} & \Rightarrow {{t}_{6}}=a{{r}^{6-1}}=72 \\\ & \Rightarrow {{t}_{6}}=a{{r}^{5}}=72.......(3) \\\ \end{aligned}$
Dividing eq. (2) by eq. (3) we get,
$\begin{aligned} & \Rightarrow \dfrac{{{t}_{8}}}{{{t}_{6}}}=\dfrac{a{{r}^{7}}}{a{{r}^{5}}}=\dfrac{60}{72} \\\ & \Rightarrow \dfrac{a{{r}^{7}}}{a{{r}^{5}}}=\dfrac{60}{72} \\\ \end{aligned}$
In the above equation, $a{{r}^{5}}$ will be cancelled out from the numerator and the denominator and we get,
$\Rightarrow {{r}^{2}}=\dfrac{60}{72}$ …….. (4)
The numerator and the denominator of the R.H.S of the above equation will get divided by 12 and we get,
$\Rightarrow {{r}^{2}}=\dfrac{5}{6}$
Now, to find the value of ${{t}_{10}}$ we are going to use the general term of the geometric sequence:
$\begin{aligned} & \Rightarrow {{t}_{10}}=a{{r}^{10-1}} \\\ & \Rightarrow {{t}_{10}}=a{{r}^{9}} \\\ \end{aligned}$
Using eq. (2) in the above equation we get,
$\begin{aligned} & \Rightarrow {{t}_{10}}=a{{r}^{7}}\left( {{r}^{2}} \right) \\\ & \Rightarrow {{t}_{10}}=60\left( {{r}^{2}} \right) \\\ \end{aligned}$
Now, substituting the value of ${{r}^{2}}$ using eq. (4) we get,
$\begin{aligned} & \Rightarrow {{t}_{10}}=60\left( \dfrac{5}{6} \right) \\\ & \Rightarrow {{t}_{10}}=50 \\\ \end{aligned}$
Hence, we have found the value of ${{t}_{10}}$ as 50.

Note: The mistake that could be possible in the above problem in writing the general term of the geometric sequence. The correct general term of the geometric sequence is as follows:
${{t}_{n}}=a{{r}^{n-1}}$
The mistake which could happen in the above problem is that you might write n in place of (n – 1) in the above formula and then the incorrect formula will be:
${{t}_{n}}=a{{r}^{n}}-1$
To remember the correct formula, the trick is that the power of r is 1 less than which is written in the subscript of t.