Question

If in an AP ,sum of the pth term is 1/q and qth term is 1/p. Prove that sum of the first(pq) terms is 1/2(pq + 1)​

Answer 1

Step-by-step explanation:

Let the first term of the given AP be 'a' and common difference be 'd',

Then,

$a_{p} = \frac{1}{q} \: and \: a_{q} = \frac{1}{p}$

$\implies \: a + (p - 1)d = \frac{1}{q} ....(i) \: \: and \: \: a + (q - 1)d = \frac{1}{p } ....(ii)$

Subtracting both the equations,

$\implies \: \{a + (p - 1)d \} - \{a + (q - 1)d \}= \frac{1}{q} - \frac{1}{p }$

$\implies \: a + (p - 1)d - a - (q - 1)d \}= \frac{p - q}{pq}$

$\implies \: \{(p - 1) - (q - 1) \}d = \frac{p - q}{pq}$

$\implies \: \{p - 1 - q + 1 \}d = \frac{p - q}{pq}$

$\implies \: (p - q )d = \frac{p - q}{pq}$

$\implies \: d = \frac{1}{pq}$

putting the value of d in equation (i),

$a + (p - 1). \frac{1}{pq} = \frac{1}{q}$

$\implies \: a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q}$

$\implies \: a = \frac{1}{pq}$

Now, sum of first pq terms is

$S _{pq} = \frac{pq}{2} .[2. \frac{1}{pq} + (pq - 1) \frac{1}{pq} ]$

$\implies \: S _{pq} = \frac{pq}{2} .[ \frac{2}{pq} + 1 - \frac{1}{pq} ]$

$\implies \: S _{pq} = \frac{pq}{2} .[ \frac{1}{pq} + 1 ]$

$\implies \: S _{pq} = \frac{(1 + pq)}{2}$