Show that the sum of $(m+n)^{ th }$ and $(m-n)^{ th }$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.
Show that the sum of $(m+n)^{ th }$ and $(m-n)^{ th }$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.
Let $a$ and $d$ be the first term and the common difference of the $A.P.$ respectively. It is known that the $k^{th}$ term of an $A.P.$ is given by
$a_{k}=a+(k-1) d$
$\therefore a_{m+n}=a+(m+n-1) d$
$a_{m-n}=a+(m-n-1) d$
$a_{m}=a+(m-1) d$
$\therefore a_{m+n}+a_{m-n}=a+(m+n-1) d+a+(m-n-1) d$
$=2 a+(m+n-1+m-n-1) d$
$=2 a+(2 m-2) d$
$=2 a+2(m-1) d$
$=2[a+(m-1) d]$
$=2 a_{m}$
Thus, the sum of $(m+n)^{t h}$ and $(m-n)^{t h}$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.
Similar Questions
If $a_1 , a_2, a_3, . . . . , a_n, ....$ are in $A.P.$ such that $a_4 - a_7 + a_{10}\, = m$, then the sum of first $13$ terms of this $A.P.$, is .............. $\mathrm{m}$
If $a_1 , a_2, a_3, . . . . , a_n, ....$ are in $A.P.$ such that $a_4 - a_7 + a_{10}\, = m$, then the sum of first $13$ terms of this $A.P.$, is .............. $\mathrm{m}$
- [JEE MAIN 2013]
The mean of the series $a,a + nd,\,\,a + 2nd$ is
The mean of the series $a,a + nd,\,\,a + 2nd$ is
If ${\log _3}2,\;{\log _3}({2^x} - 5)$ and ${\log _3}\left( {{2^x} - \frac{7}{2}} \right)$ are in $A.P.$, then $x$ is equal to
If ${\log _3}2,\;{\log _3}({2^x} - 5)$ and ${\log _3}\left( {{2^x} - \frac{7}{2}} \right)$ are in $A.P.$, then $x$ is equal to
- [IIT 1990]
If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of last and first term is $273$, then the numbers are
If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of last and first term is $273$, then the numbers are
In an $\mathrm{A.P.}$ if $m^{\text {th }}$ term is $n$ and the $n^{\text {th }}$ term is $m,$ where $m \neq n$, find the ${p^{th}}$ term.
In an $\mathrm{A.P.}$ if $m^{\text {th }}$ term is $n$ and the $n^{\text {th }}$ term is $m,$ where $m \neq n$, find the ${p^{th}}$ term.