Question

9. If p th term of an A.P. is 1/q and qth term is 1/p, prove that the sum
of the first pq terms is 1/2 [pq+1]
Hint : First find the values of a and d. Use the formula :

tn = a + (n − 1)d (ii) Sn = {2a + (n − 1) d)​

Answer 1

Solution :-

p th term of an AP = 1/q

Using n th term of an AP formula

$t_n = a + (n - 1)d$

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

$\implies a + pd - d = \dfrac{1}{q}$

$\implies q(a + pd - d) = 1$

$\implies aq + pqd - qd = 1 - - - eq(1)$

q that term of an AP = 1/p

Using n th term of an AP formula

$t_n = a + (n - 1)d$

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

$\implies a + qd - d= \dfrac{1}{p}$

$\implies p(a + qd - d)= 1$

$\implies ap+ pqd - pd= 1 - - - eq(2)$

From eq(1) and eq(2)

⇒ aq + pqd - qd = ap + pqd - pd

⇒ aq - qd = ap - pd

⇒ aq - ap = qd - pd

⇒ a(q - p) = d(q - p)

⇒ a/d = (q - p) /(q - p)

⇒ a/d = 1

⇒ a = d

Substituting a = d in eq(2)

⇒ ap + pqd - pd = 1

⇒ pd + pqd - pd = 1

⇒ pqd = 1

Substituting d = a

⇒ pqa = 1

⇒ a = 1/pq

Sum of first pq terms = Sₙ = n/2 [2a + (n - 1)d]

$\implies S_{pq} = \dfrac{pq}{2} \bigg(2 ( \dfrac{1}{pq}) + (pq - 1)d \bigg)$

$\implies S_{pq} = \dfrac{pq}{2} \bigg(\dfrac{2}{pq} + (pq - 1)d \bigg)$

$\implies S_{pq} = \dfrac{pq}{2} \bigg(\dfrac{2}{pq} + (pq - 1)a \bigg)$

[ Because a = d ]

$\implies S_{pq} = \dfrac{pq}{2} \bigg(\dfrac{2}{pq} + (pq - 1)( \dfrac{1}{pq}) \bigg)$

[ Because a = 1/pq ]

$\implies S_{pq} = \dfrac{pq}{2} \bigg(\dfrac{2}{pq} + \dfrac{pq - 1}{pq} \bigg)$

$\implies S_{pq} = \dfrac{pq}{2} \bigg(\dfrac{2 + pq - 1}{pq} \bigg)$

$\implies S_{pq} = \dfrac{ \cancel{pq}}{2} \bigg(\dfrac{ pq + 1 }{ \cancel{pq}} \bigg)$

$\implies S_{pq} = \dfrac{1}{2} (pq + 1)$

Hence proved