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If the $p^{th}$ term of an arithmetic progression is $\alpha$ and the $q^{th}$ term is $\beta$, prove that the sum of its $p+q$ term is $$\begin{align}\frac{p+q}{2}\frac{(\alpha+\beta)+(\alpha-\beta)}{p-q}\end{align}.$$

Here can we take the $p^{th}$ and $q^{th}$ term as consecutive terms of the arithmetic progression?

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do you mean consecutive? –  lab bhattacharjee Dec 18 '12 at 11:12
    
Yes, It's consecutive. Sorry! –  Alpha Dec 18 '12 at 12:44
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4 Answers

Let the arithmetic progression be $\langle x_1,x_2,x_3,\dots\rangle$, so that $\alpha=x_p$ and $\beta=x_q$. Let $$S=x_1+x_2+\ldots+x_{p+q}\;,$$ the sum of the first $p+q$ terms. Let $d$ be the constant difference of this progression, so that $x_{k+1}=x_k+d$ for every $k$. Then $x_{q+1}=x_q+d$, $x_{q+2}=x_q+2d$, and in general $x_{q+k}=x_q+kd$. In particular, if we set $k=p-q$, then $x_p=x_{q+k}=x_q+kd$, i.e., $\alpha=\beta+(p-q)d$, and it follows that $$d=\frac{\alpha-\beta}{p-q}\;.$$

Now write out the sum $S$ twice, as shown below, and add:

$$\begin{array}{c} S&=&x_1&+&x_2&+&\ldots&+&x_{p+q-1}&+&x_{p+q}\\ S&=&x_{p+q}&+&x_{p+q-1}&+&\ldots&+&x_2&+&x_1\\ \hline \end{array}$$

On the left you get $2S$. Each column on the right has the form: it contains $x_k$ and $x_{p+q+1-k}$, and its sum is $x_k+x_{p+q+1-k}$. In particular, for $k=p$ we get the sum $x_p+x_{q+1}=\alpha+\beta+d$.

Each column has the same sum (why?), and there are $p+q$ columns, so

$$2S=(p+q)(\alpha+\beta+d)\;.$$

If you now put all of the pieces together properly, you’ll get the desired formula.

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If $a,d$ be the 1st term & the common difference respectively,

$\alpha=a+(p-1)d$ and $\beta=a+(q-1)d$

Solve for $a,d$

The sum will be $\frac{p+q}2\{2a+(p+q-1)d\}$

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Ok, so i have to consider the $p^{th}$ and $q^{th}$ terms as the consecutive term. Right? –  Alpha Dec 21 '12 at 16:57
    
@Alpha, do you mean $q=p+1$ or $p=q+1?$ –  lab bhattacharjee Dec 21 '12 at 17:00
    
I meant that are $p^{th}$ and $q^{th}$ term are consecutive term of an AP –  Alpha Dec 21 '12 at 17:14
    
@Alpha, no they can be any two term of the A.P. –  lab bhattacharjee Dec 21 '12 at 17:22
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$$x_p=x_1+(p-1)d=\alpha$$ $$x_q=x_1+(q-1)d=\beta$$

$$x_1+(p-1)d-(x_1+(q-1)d)=\alpha-\beta$$

$$pd-d-qd+d=\alpha-\beta$$

$$d=\frac{\alpha-\beta}{p-q}$$ $$x_1=\alpha-(p-1)d$$ $$S_{p+q}=\frac{p+q}{2}(2x_1+(p+q-1)d)$$

$$S_{p+q}=\frac{p+q}{2}\left(2(\alpha-(p-1)\frac{\alpha-\beta}{p-q})+(p+q-1)\frac{\alpha-\beta}{p-q}\right)$$

$$S_{p+q}=\frac{p+q}{2}\left(\frac{\alpha-\beta}{p-q}(2\alpha-p+q+1)\right)$$

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up vote 0 down vote accepted

So here I have to consider the $p^{th}$ and $q^{th}$ as the consecutive term so that by doing their difference I can find the Common difference of the whole AP. So,

$a_p=\alpha=a+(p-1)d$   ------- (i)

and

$a_q=\beta=a+(q-1)d$   -------- (ii)

By substracting (i) and (ii) we get,

$[a+(p-1)d]-[a+(q-1)d]$

After solving them with the sum formula of the AP, $S_n= \frac n2[2a+(n-1)d]$, we can proceed by $\frac{p+q}2\{2a+(p+q-1)d\}$ and can prove that they are equal.

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