Sequence And Series Question 84
Question: Consider the ten numbers $ ar,ar^{2},ar^{3},…ar^{10} $ .If their sum is 18 and the sum the reciprocals is 6, then the product of these ten numbers is
Options:
A) 81
B) 243
C) 343
D) 324
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given $ \frac{ar(r^{10}-1)}{r-1}=18 $ …(1) Also $ \frac{\frac{1}{ar}( 1-\frac{1}{r^{10}} )}{1-\frac{1}{r}}=6 $ Or $ \frac{1}{ar^{11}}.\frac{(r^{10}-1)r}{r-1}=6 $ Or $ \frac{1}{a^{2}r^{11}}.\frac{ar(r^{10}-1)}{r-1}=6 $ …(2) From (1) and (2) $ \frac{1}{a^{2}r^{11}}.\times 18=6 $ Or $ a^{2}r^{11}=3 $ Now $ P=a^{10}r^{11}={{(a^{2}r^{11})}^{5}}=3^{5}=243 $
BETA
