Question

If the $m^{th}$ term of an A.P. is $\frac{1}{n}$ and the $n^{th}$ term is $\frac{1}{m}$ , show that the sum of mn terms is ($\frac{1}{2}$)(mn + 1).

Answer 1

Let a be the first term and d be the common difference of the given A.P. Then ,

$a_m$ = $\frac{1}{n}$

a + (m - 1)d = $\frac{1}{n}$ ... (i)

and

$a_n$ = $\frac{1}{m}$

a + (n - 1)d= $\frac{1}{m}$ ... (ii)

Subtracting equation (ii) from equation (i), we get

(m - n)d = $\frac{1}{n}$ - $\frac{1}{m}$

(m - n)d = $\frac{(m-n)}{mn}$

d = $\frac{1}{mn}$

Putting d = $\frac{1}{mn}$ in equation (i), we get

a + (m - 1)$\frac{1}{mn}$ = $\frac{1}{n}$

a + $\frac{1}{n}$ - $\frac{1}{mn}$ = $\frac{1}{n}$

a = $\frac{1}{mn}$

Now , $S_{mn}$ = ($\frac{mn}{2}$) {2a + (mn - 1)d}

Smn = ($\frac{mn}{2}$) {($\frac{2}{mn}$) + (mn - 1)d}

$S_{mn}$< = ($\frac{1}{2}$) (mn + 1)