A) \[\alpha -\beta \]
B) \[\beta -\alpha \]
C) \[\frac{\alpha -\beta }{2}\]
D) None
Correct Answer: D
Solution :
[d] Let d be the common difference of the A.P. Then \[{{a}_{2r}}={{a}_{2r-1}}+d\]. \[\therefore \sum\limits_{r=1}^{100}{{{a}_{2r}}=\sum\limits_{r=1}^{100}{({{a}_{2r-1}}+d)}=\sum\limits_{r=1}^{100}{{{a}_{2r-1}}+100d}}\] \[\Rightarrow \alpha =\beta +100\,d\Rightarrow d=\frac{\alpha -\beta }{100}\]
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