11th Class Mathematics Other Series Question Bank Critical Thinking

  • 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

    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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