Question

If the pth term of an AP is 1/q and with term is 1/p prove that the sum of the first pq terms of the AP is pq+1÷2

Answer 1

Question: If the pth term of an AP is 1/q and the qth term of an AP is 1/p, Prove that the Sum of the first pq terms of the AP is pq + 1/2

ATQ,

$\sf A_p = \dfrac{1}{q}$

$\sf A_q = \dfrac{1}{p}$

To Prove that,

$\sf S_{pq} = \dfrac{pq + 1}{2}$

Solution:

$\sf A_p = \dfrac{1}{q}$

$\sf a + (p-1)d = \dfrac{1}{q}$

Let the above be Equation 1.

We know that,

$\sf A_q = \dfrac{1}{p}$

$\sf a + (q-1)d = \dfrac{1}{p}$

Let this be Equation 2.

Subtracting Equation 2 from Equation 1 we get,

(Division given in attachment)

$\sf a + (p-1)d = \dfrac{1}{q}\\ \\ a + (q-1)d = \dfrac{1}{p}}\\ \\ \rule{85}{1}$

$\Longrightarrow \sf (p - 1)d - [(q - 1)d] = \dfrac{1}{q} - \dfrac{1}{p}$

$\Longrightarrow \sf \ pd - d - [qd - d] = \dfrac{p-q}{pq}$

$\Longrightarrow \ \sf pd - d - qd + d = \dfrac{p-q}{pq}$

$\Longrightarrow \ \sf pd - qd = \dfrac{p-q}{pq}$

$\Longrightarrow \ \sf d(p - q) = \dfrac{p-q}{pq}$

$\Longrightarrow \ \sf d = \dfrac{p-q}{(p-q)pq}$

$\Longrightarrow \ \sf d = \dfrac{1}{pq}$

$\implies \boxed{\sf d = \frac{1}{pq}}$

Now, we Substitute the value of 'd' in Equation 1.

$\Longrightarrow \ \sf a + (p-1)d = \dfrac{1}{q}$

$\Longrightarrow \ \sf a + (p-1)\dfrac{1}{pq} = \dfrac{1}{q}$

$\Longrightarrow \ \sf a + \dfrac{(p-1)}{pq} = \dfrac{1}{q}$

$\Longrightarrow \ \sf a = \dfrac{1}{q} - \dfrac{(p-1)}{pq}$

$\Longrightarrow \ \sf a = \dfrac{p-(p-1)}{pq}$

$\Longrightarrow \ \sf a = \dfrac{p-p+1}{pq}$

$\Longrightarrow \ \sf a = \dfrac{1}{pq}$

$\implies \boxed{\sf a = \frac{1}{pq}}$

Now, We find the Sum of the first pq terms of the AP.

Sum of First pq terms of this AP is $\sf S_{pq}$

$\Longrightarrow \sf S_{pq} = \dfrac{n}{2} \left( 2a + (n-1)d \right)$

$\Longrightarrow \sf S_{pq} = \dfrac{pq}{2} \left(2\left(\frac{1}{pq}\right) + (pq-1)\frac{1}{pq}\right)$

$\Longrightarrow \sf S_{pq} = \dfrac{pq}{2} \left(\frac{2}{pq} + \frac{(pq-1)}{pq}\right)$

$\Longrightarrow \sf \ S_{pq} = \dfrac{pq}{2} \left(\dfrac{2+pq-1}{pq}\right)$

$\Longrightarrow \sf \ S_{pq} = \dfrac{pq}{2} \left(\dfrac{1+pq}{pq}\right)$

['Pq' is cancelled]

$\Longrightarrow \sf \ S_{pq} = \left(\dfrac{pq+1}{2}\right)$

Hence Proved.