Sequence And Series Question 338
Question: Let $ a_{n} $ be the nth term of an A.P. If $ \sum\limits_{r=1}^{100}{a_{2r}=\alpha } $ and $ \sum\limits_{r=1}^{100}{{a_{2r-1}}=\beta } $ , then the common difference of the A.P. is
Options:
A) $ \alpha -\beta $
B) $ \beta -\alpha $
C) $ \frac{\alpha -\beta }{2} $
D) None
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let d be the common difference of the A.P. Then $ a_{2r}={a_{2r-1}}+d $ .
$ \therefore \sum\limits_{r=1}^{100}{a_{2r}=\sum\limits_{r=1}^{100}{({a_{2r-1}}+d)}=\sum\limits_{r=1}^{100}{{a_{2r-1}}+100d}} $
$ \Rightarrow \alpha =\beta +100,d\Rightarrow d=\frac{\alpha -\beta }{100} $
BETA
