If the sum of first m terms of an AP is same as the sum of its first n terms (m not equal to n ), show that the sum of its first (m+n) terms is zero.
Answers
-------(1)
Answer:
\underline{\textbf{Given:}}
Given:
\textsf{Sum of first m terms of an A.P is same as the sum of its first n terms}Sum of first m terms of an A.P is same as the sum of its first n terms
\underline{\textbf{To find:}}
To find:
\textsf{The sum of its first (m+n) terms is zero}The sum of its first (m+n) terms is zero
\underline{\textbf{Solution:}}
Solution:
\underline{\textbf{Concept used:}}
Concept used:
\textsf{The sum of first n terms of an A.P}The sum of first n terms of an A.P
\mathsf{a,a+d,a+2d,\;.\;.\;.\;.\;.\;.is}a,a+d,a+2d,......is
\boxed{\mathsf{S_n=\dfrac{n}{2}[2a+(n-1)d]}}
S
n
=
2
n
[2a+(n−1)d]
\textsf{As per given data,}As per given data,
\mathsf{S_m=S_n}S
m
=S
n
\mathsf{\dfrac{m}{2}[2a+(m-1)d]=\dfrac{n}{2}[2a+(n-1)d]}
2
m
[2a+(m−1)d]=
2
n
[2a+(n−1)d]
\mathsf{m[2a+(m-1)d]=n[2a+(n-1)d]}m[2a+(m−1)d]=n[2a+(n−1)d]
\mathsf{2am+m^2d-md=2an+n^2-nd}2am+m
2
d−md=2an+n
2
−nd
\textsf{Rearranging terms, we get}Rearranging terms, we get
\mathsf{2a(m-n)+(m^2-n^2)d=(m-n)d}2a(m−n)+(m
2
−n
2
)d=(m−n)d
\mathsf{2a(m-n)+(m-n)(m+n)d=(m-n)d}2a(m−n)+(m−n)(m+n)d=(m−n)d
\textsf{Divide bothsides by m-n}Divide bothsides by m-n
\mathsf{2a+(m+n)d=d}2a+(m+n)d=d
\mathsf{2a+(m+n)d-d=0}2a+(m+n)d−d=0
\mathsf{2a+((m+n)-1)d=0}2a+((m+n)−1)d=0 -------(1)
\mathsf{Now,}Now,
\mathsf{S_{m+n}=\dfrac{m+n}{2}[2a+((m+n)-1)d]}S
m+n
=
2
m+n
[2a+((m+n)−1)d]
\mathsf{S_{m+n}=\dfrac{m+n}{2}[0]}\;\;\;\;\;\textbf{(Using (1))}S
m+n
=
2
m+n
[0](Using (1))
\implies\boxed{\mathsf{S_{m+n}=0}}⟹
S
m+n
=0
\textsf{Hence proved}Hence proved