• # question_answer If ${{a}_{1}},\ {{a}_{2}},\,{{a}_{3}},......{{a}_{24}}$ are in arithmetic progression and ${{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{24}}=225$, then ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+........+{{a}_{23}}+{{a}_{24}}=$ [MP PET 1999; AMU 1997] A) 909 B) 75 C) 750 D) 900

${{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{24}}=225$ $\Rightarrow$ $({{a}_{1}}+{{a}_{24}})+({{a}_{5}}+{{a}_{20}})+({{a}_{10}}+{{a}_{15}})=225$ $\Rightarrow$ $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) ${{a}_{1}}+{{a}_{2}}+......+{{a}_{24}}=\frac{24}{2}({{a}_{1}}+{{a}_{24}})=12\times 75=900$.