question_answer

The sums of first n terms of three arithmetic progressions are \[{{S}_{1}},{{S}_{2}}\] and \[{{S}_{3}}\] respectively. The first term of each A.P. is 1 and their common differences are 1, 2 and 3 respectively. Prove that \[{{S}_{1}}+{{S}_{3}}=2{{S}_{2}}\].

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.