Sequence And Series Question 235
Question: If $ n $ geometric means be inserted between $ a $ and $ b $ then the $ n^{th} $ geometric mean will be
Options:
A) $ a,{{( \frac{b}{a} )}^{\frac{n}{n-1}}} $
B) $ a,{{( \frac{b}{a} )}^{\frac{n-1}{n}}} $
C) $ a,{{( \frac{b}{a} )}^{\frac{n}{n+1}}} $
D) $ a,{{( \frac{b}{a} )}^{\frac{1}{n}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
If $ n $ geometric means $ g_1,g_2…….g_{n} $ are to be inserted between two positive real numbers $ a $ and $ b $ , then $ a,\ g_1,\ g_2……g_{n},\ b $ are in G.P. Then $ g_1=ar,\ g_2=ar^{2}……..g_{n}=ar^{n} $ So $ b=a{r^{n+1}}\Rightarrow r={{( \frac{b}{a} )}^{1/(n+1)}} $ Now $ n^{th} $ geometric mean $ $ $ (g_{n})=ar^{n}=a{{( \frac{b}{a} )}^{n/(n+1)}} $ . Aliter : As we have the $ m^{th} $ G.M. is given by $ G_{m}=a{{( \frac{b}{a} )}^{\frac{m}{n+1}}} $ Now replace $ m $ by $ n $ we get the required result.
BETA
