Question

In an ap if the pth term is 1/q and the qth term is 1/p prove that the sum of first pq terms is 1+pq/2 where p is not equal to q

Answer 1

STEP-BY-STEP EXPLANATION :-

$a + (p - 1)d = \frac{1}{q}.....(i) \\ \\ a + (q - 1)d = \frac{1}{p} .......(ii) \\ \\ \\ \\ subtracting \: them \: \\ \\ \\ a + (p - 1)d - a - (q - 1)d = \frac{p - q}{pq} \\ \\ \\ d(p - 1 - q + 1) = \frac{p - q}{pq} \\ \\ d(p - q) = \frac{p - q}{pq} \\ \\ d = \frac{1}{pq} \\ \\ \\ putting \: in \: (i) \\ \\ a + (p - 1) \frac{1}{pq} = \frac{1}{q} \\ \\ a + \frac{p}{pq} - \frac{1}{pq} = \frac{1}{q} \\ \\ a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q} \\ \\ a = \frac{1}{pq}$

Now,

We have a as well as d.

So,

Sum of first pq terms will be :-

$s = \frac{pq}{2} (2a + (pq - 1)d) \\ \\ = \frac{pq}{2} (2 \times \frac{1}{pq} + (pq - 1) \frac{1}{pq} ) \\ \\ \\ = \frac{pq}{2} ( \frac{2 + pq - 1}{pq} ) \\ \\ \\ \frac{1}{2} (pq + 1)$

Hence,

Proved!

Answer 2

$Given:-$

• $pth \: \: term = \: \frac{1}{q}$

• $qth \: \: term = \: \frac{1}{p}$

$To \: Prove:-$

• $Sum \: \: of \: \: first \: \: pq \: \: terms \: \: is \: \: \frac{1}{2}(pq + 1)$

$Solutions:-$

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

• $pth \: \: term \: \: is \: \: \frac{1}{q}$

$⇢ a_p = a + (p - 1)d$

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

• $qth \: \: term \: \: is \: \: \frac{1}{p}$

$⇢ a_q = a + (q - 1)d$

$⇢ a + (q - 1)d = \frac{1}{p}$......(2).

• $Substracting \: \: Eq. \: \: (2). \: \: from \: \: (1).$

$⇢ a + (p - 1)d = \frac{1}{q}$

$⇢ a + (q - 1)d = \frac{1}{p}$

$\: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: = - \: \: \: \: \: \: \: \: \: \: \: \:$ _____________________

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

$⇢ pd - d \: - \: qd + d = \frac{p - q}{pq}$

$⇢ pd \: - \: qd = \frac{p - q}{pq}$

$⇢ (p - q)d = \frac{p - q}{pq}$

$⇢ d = \frac{p - q}{pq} × \frac{1}{p - q}$

$⇢ d = \frac{1}{pq}$

• $putting \: \: value \: \: of \: \: d = \: \: \frac{1}{pq} \: \: in \: \: Eq. \: \: (1).$

$⇢ a + (p - 1)d = \frac{1}{q}$

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

$⇢ a + \frac{p}{pq} - \frac{1}{pq} = \frac{1}{q}$

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

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

$⇢ a = \frac{1}{pq}$

• $So, \: \: a \: \: = \: \: \frac{1}{pq} \: \: \: and \: \: \: d= \: \: \frac{1}{pq}$

$S_n = \frac{n}{2}[2a × (n - 1)d]$

• $Sum \: \: of \: \: 1st \: \: pq \: \: terms:-$

$⇢ S_pq = \frac{pq}{2}[2 × \frac{1}{pq} × (pq - 1) × \frac{1}{pq}]$

$⇢ S_pq = \frac{pq}{2}[ \frac{2}{pq} + \frac{pq}{pq} + \frac{1}{pq}]$

$⇢ S_pq = \frac{pq}{2}[ \frac{2 + pq - 1}{pq}]$

$⇢ S_pq = \frac{pq}{2}[ \frac{1 + pq}{pq}]$

$⇢ S_pq = \frac{1}{2}[pq + 1]$

$hence, \: \: \: proved \: \: that \: \: the \: \: sum \: \: of \: \: first \: \: pq \: \: therms \: \: is \: \: = \: \: \frac{1}{2}(pq + 1), \: \: where \: \: p \: \: is \: \: not \: \: equal \: \: to \: \: q.$

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