Sequence And Series Question 176
Question: If $ n $ geometric means between $ a $ and $ b $ be $ G_1,\ G_2,\ ….. $ $ G_{n} $ and a geometric mean be $ G $ , then the true relation is
Options:
A) $ G_1.G_2……..G_{n}=G $
B) $ G_1.G_2……..G_{n}={G^{1/n}} $
C) $ G_1.G_2……..G_{n}=G^{n} $
D) $ G_1.G_2……..G_{n}={G^{2/n}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Here $ G={{(ab)}^{1/2}} $ and $ G_1=ar^{1},\ G_2=ar^{2},……..G_{n}=ar^{n} $ Therefore $ G_1.\ G_2.\ G_3…..G_{n}=a^{n}{r^{1+2+…+n}}=a^{n}{r^{n(n+1)/2}} $ But $ a{r^{n+1}}=b\Rightarrow r={{( \frac{b}{a} )}^{1/(n+1)}} $ Therefore, the required product is $ a^{n}{{( \frac{b}{a} )}^{1/(n+1)\ .\ n(n+1)/2}}={{(ab)}^{n/2}}={{{ {{(ab)}^{1/2}} }}^{n}}=G^{n} $ . Note: It is a well-known fact.
BETA
