11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer The sums of \[n\] terms of three A.P.'s whose first term is 1 and common differences are 1, 2, 3 are \[{{S}_{1}},\ {{S}_{2}},\ {{S}_{3}}\] respectively. The true relation is

    A) \[{{S}_{1}}+{{S}_{3}}={{S}_{2}}\]

    B) \[{{S}_{1}}+{{S}_{3}}=2{{S}_{2}}\]

    C) \[{{S}_{1}}+{{S}_{2}}=2{{S}_{3}}\]

    D) \[{{S}_{1}}+{{S}_{2}}={{S}_{3}}\]

    Correct Answer: B

    Solution :

    We have \[{{a}_{1}}={{a}_{2}}={{a}_{3}}=1\]and \[{{d}_{1}}=1,\ {{d}_{2}}=2,\ {{d}_{3}}=3\]. Therefore,  \[{{S}_{1}}=\frac{n}{2}(n+1)\]                ......(i)                    \[{{S}_{2}}=\frac{n}{2}[2n]\]               ......(ii)                    \[{{S}_{3}}=\frac{n}{2}[3n-1]\]              ......(iii) Adding (i) and (iii), \[{{S}_{1}}+{{S}_{3}}=\frac{n}{2}[(n+1)+(3n-1)]=2\left[ \frac{n}{2}(2n) \right]=2{{S}_{2}}\] Hence correct relation\[{{S}_{1}}+{{S}_{3}}=2{{S}_{2}}\].

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