• # 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) ${{S}_{1}}+{{S}_{2}}={{S}_{3}}$

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${{S}_{1}}+{{S}_{3}}=2{{S}_{2}}$.