if the pth term of an A P is 1/q and the qth term is 1/p,show that the sum - Maths - Quadratic Equations

Question

if the pth term of an A P is 1/q and the qth term is 1/p,show
that the sum of pq terms is 1/2(pq+1)

Lalit Mehra · Accepted Answer

Dear Student!

Let a and d be the first term and common difference of A.P. respectively.

$n^{t h} term of the A.P., a_{n} = a + (n - 1) d Given, a_{p} = \frac{1}{q} and a_{q} = \frac{1}{p} a_{p} = \frac{1}{q} ∴ a + (p - 1) d = \frac{1}{q} ... (1) a_{q} = \frac{1}{p} ∴ a + (q - 1) d = \frac{1}{p} ... (2)$

Solving (1) and 2, we get

$a + (p - 1) d = \frac{1}{q} a + {(q - 1) d = \frac{1}{p}}^{} \underset{̲}{- - -} [(p - 1) - (q - 1)] d = \frac{1}{q} - \frac{1}{p} ∴ (p - q) d = \frac{(p - q)}{p q} \Rightarrow d = \frac{1}{p q} W hen d = \frac{1}{p q}, we get a + (p - 1) \times \frac{1}{p q} = \frac{1}{q} (U \sin g (1)) ∴ a = \frac{1}{q} - (p - 1) \times \frac{1}{p q} = \frac{1}{q} - \frac{1}{q} + \frac{1}{p q} = \frac{1}{p q}$

Sum of pq terms of the A.P. = S_pq

$S_{p q} = \frac{p q}{2} [2 \times \frac{1}{p q} + (p q - 1) \times \frac{1}{p q}] (∵ S_{n} = \frac{n}{2} [2 a + (n - 1) d]) ∴ S_{p q} = \frac{p q}{2} (\frac{2}{p q} + 1 - \frac{1}{p q}) \Rightarrow S_{p q} = \frac{p q}{2} (\frac{1}{p q} + 1) \Rightarrow S_{p q} = \frac{1}{2} {(1 + p q)}^{} = \frac{1}{2} (p q + 1)$

Thus, the sum of pq terms of the given A.P. is $\frac{1}{2} (p q + 1)$ .

Cheers!

if the pth term of an A.P. is 1/q and the qth term is 1/p,show that the sum of pq terms is 1/2(pq+1).