Sequence And Series Question 420
Question: Let a = 111… 1(55 digits), $ b=1+10+10^{2}+…+10^{4}, $ $ c=1+10^{5}+10^{10}+10^{15}+….+10^{50}, $ then
Options:
A) $ a=b+c $
B) $ a=bc $
C) $ b=ac $
D) $ c=ab $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ a=1+10+10^{2}+….+10^{54} $ $ =\frac{10^{55}-1}{10-1}=\frac{10^{55}-1}{10^{5}-1}.\frac{10^{5}-1}{10-1}=bc $
BETA
