Question

Let $ u,v,w $ be real numbers in geometric progression such that $ u>v>w $ . Suppose $ {{u}^{40}}={{v}^{n}}={{w}^{60}} $ . Find the value of n.

Answer 1

Hint: From the given numbers of geometric sequence, we find the general term of the series. We find the formula for $ {{t}_{n}} $ , the $ {{n}^{th}} $ term of the series. From the given sequence we find the condition of $ uw={{v}^{2}} $ . Then we put the value of consecutive natural numbers for $ n $ to find the solution.

Complete step by step solution:
We have been given three numbers of geometric sequences which are $ u,v,w $ .
We express the geometric sequence in its general form.
We express the terms as $ {{t}_{n}} $ , the $ {{n}^{th}} $ term of the series.
The first term be $ {{t}_{1}} $ and the common ratio be $ r $ where $ r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}} $ .
We can express the general term $ {{t}_{n}} $ based on the first term and the common ratio.
The formula being $ {{t}_{n}}={{t}_{1}}{{r}^{n-1}} $ .
Now if three numbers \[x,y,z\] are in geometric progression then we can say that $ xz={{y}^{2}} $ .
Putting the values, we get $ uw={{v}^{2}} $ . We can also assume that $ {{w}^{20}}={{v}^{\dfrac{n}{3}}} $ .
It’s also given that $ {{u}^{40}}={{v}^{n}}={{w}^{60}} $ .
Therefore, $ {{u}^{40}}\times {{w}^{60}}={{\left( {{v}^{n}} \right)}^{2}}={{v}^{2n}} $ .
 $ \begin{align}
  & {{u}^{40}}\times {{w}^{60}}={{v}^{2n}} \\
 & \Rightarrow {{u}^{40}}\times {{w}^{40}}\times {{w}^{20}}={{v}^{2n}} \\
 & \Rightarrow {{\left( uw \right)}^{40}}\times {{v}^{\dfrac{n}{3}}}={{v}^{2n}} \\
 & \Rightarrow {{v}^{80}}\times {{v}^{\dfrac{n}{3}}}={{v}^{2n}} \\
 & \Rightarrow {{v}^{80+\dfrac{n}{3}}}={{v}^{2n}} \\
\end{align} $
Equating the indices, we get $ 80+\dfrac{n}{3}=2n $ .
Solving we get
 $ \begin{align}
  & 80+\dfrac{n}{3}=2n \\
 & \Rightarrow \dfrac{5n}{3}=80 \\
 & \Rightarrow n=\dfrac{80\times 3}{5}=48 \\
\end{align} $
The value of n is 48.
So, the correct answer is “48”.

Note: The sequence is an increasing sequence where the common ratio is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number. The ratio formula should always be according $ r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}} $ .