Question

sum of the series a-(a+d)+(a+2d)-(a+3d)+......to(2n+1) terms is equal to​

Answer 1

We're asked to find the sum of terms of the series,

$\small\displaystyle\longrightarrow S=a-(a+d)+(a+2d)-(a+3d)+\,\dots\dots\,(2n+1)\,terms$

The n'th term of this series is,

$\small\displaystyle\longrightarrow a_n=(-1)^{n+1}(a+(n-1)d)$

Then (2n + 1)th term is,

$\small\displaystyle\longrightarrow a_{2n+1}=a+2nd$

and (2n)th term is,

$\small\displaystyle\longrightarrow a_{2n}=-(a+(2n-1)d)$

Let us define two series,

$\small\displaystyle\longrightarrow S_1=a+(a+2d)+(a+4d)+\,\dots\,+(a+2nd)$

$\small\displaystyle\longrightarrow S_2=(a+d)+(a+3d)+(a+5d)+\,\dots\,+(a+(2n-1)d)$

so that $\small\displaystyle S=S_1-S_2.$

The series $\small\displaystyle S_1$ is an AP of first term a, common difference 2d and last term (a + 2nd).

No. of terms in $\small\displaystyle S_1,$

$\small\displaystyle\longrightarrow n_1=\dfrac{(a+2nd)-a}{2d}+1$

$\small\displaystyle\longrightarrow n_1=n+1$

Then sum of terms of $\small\displaystyle S_1,$

$\small\displaystyle\longrightarrow S_1=\dfrac{n+1}{2}(a+(a+2nd))$

$\small\displaystyle\longrightarrow S_1=(n+1)(a+nd)$

The series $\small\displaystyle S_2$ is an AP of first term (a+d), common difference 2d and last term (a + (2n - 1)d).

No. of terms in $\small\displaystyle S_2,$

$\small\displaystyle\longrightarrow n_2=\dfrac{(a+(2n-1)d)-(a+d)}{2d}+1$

$\small\displaystyle\longrightarrow n_2=n$

Then sum of terms of $\small\displaystyle S_2,$

$\small\displaystyle\longrightarrow S_2=\dfrac{n}{2}((a+d)+(a+(2n-1)d))$

$\small\displaystyle\longrightarrow S_2=n(a+nd)$

Then our series is,

$\small\displaystyle\longrightarrow S=S_1-S_2$

$\small\displaystyle\longrightarrow S=(n+1)(a+nd)-n(a+nd)$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{S=a+nd}} }$