Answer:
Given, first term of each \[A.P(a)=1\]. and their common differences are 1, 2 and 3. \[\therefore {{S}_{1}}=\frac{n}{2}[2a+(n-1){{d}_{1}}]\] \[=\frac{n}{2}(2+(n-1)1)=\frac{n}{2}(n+1)\] \[{{S}_{2}}=\frac{n}{2}[2a+(n-1){{d}_{2}}]\] \[=\frac{n}{2}(2+(n-1)2)=\frac{n}{2}(2n)={{n}^{2}}\] and \[{{S}_{3}}=\frac{n}{2}[2a+(n-1){{d}_{3}}]\] \[=\frac{n}{2}(2+(n-1)3)=\frac{n}{2}(3n-1)\] Now, \[{{S}_{1}}+{{S}_{3}}=\frac{n}{2}(n+1)+\frac{n}{2}(3n-1)\] \[=\frac{n}{2}(n+1+3n-1)=4n\times \frac{n}{2}=2{{n}^{2}}\] \[=2{{S}_{2}}\] \[\therefore {{S}_{1}}+{{S}_{3}}=2{{S}_{2}}\] Hence Proved.
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