Question

Which term of the GP $2,2\sqrt{2},4,..$ is 128?

Answer 1

Hint: Calculate the common ratio of the given GP by dividing any two consecutive terms of the GP. Use the fact that the ${{n}^{th}}$ term of the GP is given by ${{a}_{n}}=a{{r}^{n-1}}$, where ${{a}_{n}}$ is the ${{n}^{th}}$ term of the GP, ‘a’ is the first term of GP and ‘r’ is the common ratio of the GP. Substitute the values in the given expression and calculate the value of ‘n’.

Complete Step-by-step answer:
We have to calculate which term of the GP $2,2\sqrt{2},4,..$ is 128.
We will first calculate the common ratio of the given GP. To do so, we will divide any two consecutive terms of the GP.
Thus, the common ratio of the given GP is $=\dfrac{2\sqrt{2}}{2}=\sqrt{2}$.
We will now calculate which term of the GP $2,2\sqrt{2},4,..$ is 128.
We know that the ${{n}^{th}}$ term of the GP is given by ${{a}_{n}}=a{{r}^{n-1}}$, where ${{a}_{n}}$ is the ${{n}^{th}}$ term of the GP, ‘a’ is the first term of GP and ‘r’ is the common ratio of the GP.
Substituting ${{a}_{n}}=128,a=2,r=\sqrt{2}$ in the above expression, we have $128=2{{\left( \sqrt{2} \right)}^{n-1}}$.
Rearranging the terms of the above equation, we have ${{\left( \sqrt{2} \right)}^{n-1}}=\dfrac{128}{2}=64$.
Thus, we can rewrite the above equation as ${{2}^{\dfrac{n-1}{2}}}={{2}^{6}}=64$.
So, we have $\dfrac{n-1}{2}=6$.
Rearranging the terms of the above equation, we have $n-1=2\times 6=12$.
Thus, we have $n=12+1=13$.
Hence, 128 is the ${{13}^{th}}$ term of the GP $2,2\sqrt{2},4,..$.

Note: We know the Geometric Progression (GP) is a sequence of numbers such that the ratio of any two consecutive numbers is a constant. We must also keep in my mind we can calculate the common ratio of the GP by dividing any two consecutive terms of the GP.