Question

The first term of a geometric sequence is 10 and the fourth term is 160. What is the common ratio?
A.1
B.2
C.3
D.4

Answer 1

Hint: Let the common ratio be $r$. We are given first term as 10. Use the formula of \[{n^{th}}\] term of geometric series. Substitute the values of first term , $n = 4$and fourth term in the formula ${a_n} = a{r^{n - 1}}$. Solve the equation to find the value of $r$.

Complete step-by-step answer:
Let us consider the common ratio of the required G.P. be $r$ and the G.P. sequence be
$a,ar,a{r^2}......$, where $a$ represents the first term of the G.P. .
Since we are given that the first term of the required G.P. is 10, and the first term of the G.P. according to our assumption is $a$ , we conclude that $a = 10$
It is known that for a geometric progression, say $a,ar,a{r^2}....$, where $a$ is the first term and $r$is the common ratio , the ${n^{th}}$ term of the G.P. can be represented by the formula ${a_n} = a{r^{n - 1}}$.
We know that the fourth term of the required G.P. is 160. Thus substituting the value 160 for ${a_4}$ and 4 for $n$in the formula ${a_n} = a{r^{n - 1}}$, we get
$
  {a_4} = a{r^3} \\
  160 = a{r^3} \\
 $
And we already concluded that $a = 10$. Substituting 10 for $a$ in the equation $160 = a{r^3}$, we get
$160 = 10{r^3}$
We can thus solve for $r$ in the equation $160 = 10{r^3}$.
$
  16 = {r^3} \\
  r = \sqrt[3]{{16}} \\
  r = 2\sqrt[3]{2} \\
 $
Therefore, the common ratio $r$ is $2\sqrt[3]{2}$.

Note: The formula for the ${n^{th}}$ term of the G.P. represented by $a,ar,a{r^2}....$, where $a$ is the first term and $r$ is the common ratio is ${a_n} = a{r^{n - 1}}$. For an even power of $r$ in the equation ${a_n} = a{r^{n - 1}}$, there can be multiple possible answers for the common ratio upon solving .