question_answer

If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first \[(m+n)\]terms is zero.

Answer:

Let a be first term and d is common difference of given A.P. then,

\[{{S}_{m}}={{S}_{n}}\]

\[\frac{m}{2}\{2a+(m-1)d\}=\frac{n}{2}\{2a+(n-1)d\}\]

\[\Rightarrow \] \[\frac{2am}{2}+\frac{m}{2}(m-1)d-\frac{2an}{2}-\frac{n}{2}(n-1)d=0\]

\[\Rightarrow \] \[2am-2an+\{m(m-1)-n(n-1)\}d=0\]

\[\Rightarrow \] \[2a(m-n)+({{m}^{2}}-m-{{n}^{2}}+n)d=0\]

\[\Rightarrow \] \[2a(m-n)+({{m}^{2}}-{{n}^{2}}-(m-n)d=0\]

\[\Rightarrow \] \[2a(m-n)+(m-n)(m+n-1)d=0\]

\[\Rightarrow \] \[(m-n)+(2a+(m+n-1)d=0\]

\[\Rightarrow \] \[2a+(m+n-1)d=0\]

Now,     \[{{S}_{m+n}}=\frac{m+n}{2}\{2a+(m+n-1)d\}\]

\[=\frac{m+n}{2}\times 0=0\]                                 Hence Proved.