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Answer 1

Solution

Let us consider the first term and common difference of the AP be a and d respectively

Given ,

$\sf p {}^{th} \: term = \dfrac{1}{q} \\ \\ \implies \sf a + (p - 1)d = \dfrac{1}{q} \dashrightarrow(1)$

And

$\sf q {}^{th} \: term = \dfrac{1}{p} \\ \\ \sf \implies a + (q - 1)d = \dfrac{1}{p} \dashrightarrow(2)$

Subtracting (1) from (2)

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

Now putting the value of ‘d’ in (1)

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

Thus ,

first term , a = 1/pq

common difference , d = 1/pq

Thus , sum upto pq terms is :

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

Where p≠q (proved)

Answer 2

Answer:

→ $\bigg (\dfrac{1}{2}\bigg ) [pq + 1]$

GIVEN :

$P^{th} term = \dfrac {1}{q}$

$q^{th} term = \dfrac {1}{p}$

TO PROVE :

The sum of first pq terms is $\dfrac {1+pq}{2}$, where $p \neq q$

SOLUTION :

→ $a + (p-1)d = \dfrac {1}{q}$

→ $aq + (pq-1)d = 1$ --------- (i)

Similarly,

→ $ap + (pq-p)d = 1$ ---------- (ii)

From equation (i) & (ii) we get,

→ $aq + (pq-q)d = ap + (pq-p)d$

→ $aq - ap = d[pq - p - pq + q]$

→ $a (q-p) = d (q-p)$

→ $a = d$

Now, eq (i) becomes,

→ $dq + pqd - dq = 1$

→ $d = \dfrac {1}{pq}$

→ $Spq = \bigg (\dfrac{pq}{2} \bigg) [2a + (pq-1)d]$

→ $\bigg (\dfrac{pq}{2} \bigg ) 2 \times \bigg (\dfrac{1}{pq} \bigg ) + (pq-1)\bigg (\dfrac{1}{pq} \bigg )$

→ $\bigg [\dfrac{1}{2} \bigg] [2 + pq - 1]$

$\Rightarrow \bigg (\dfrac{1}{2}\bigg ) [pq+1]$.

ADDITIONAL INFORMATION RELATED TO AP :

• Some formulas of ap :-

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

2. $S_n =\dfrac {n}{2} [ 2a + (n-1) d ]$

3. $S_n = \dfrac {n}{2} (a + n)$

4. $S = \dfrac {n+1}{2}$