A) 909
B) 75
C) 750
D) 900
Correct Answer: D
Solution :
\[{{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{24}}=225\] \[\Rightarrow \] \[({{a}_{1}}+{{a}_{24}})+({{a}_{5}}+{{a}_{20}})+({{a}_{10}}+{{a}_{15}})=225\] \[\Rightarrow \] \[3({{a}_{1}}+{{a}_{24}})=225\]\[\Rightarrow \]\[{{a}_{1}}+{{a}_{24}}=75\] (\[\because \] In an A.P. the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term) \[{{a}_{1}}+{{a}_{2}}+......+{{a}_{24}}=\frac{24}{2}({{a}_{1}}+{{a}_{24}})=12\times 75=900\].
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