Question

If the sum of first (p terms) of an AP is q and if the sum of the first q term of an $\text{AP}=\text{p}$. Show that
Sum $\left( \text{p}+\text{q} \right)$ term is -$\left[ \left( \text{p}+\text{q} \right) \right]$

Answer 1

Hint: Use the “formulae” for the first n term of an AP given by
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Where a is the first term of AP
d is the common difference
Try to find a relation between the sum of first p, q and $\left( \text{p}+\text{q} \right)$ terms and get the result.

Complete step-by-step answer:
As given in the question the sum of first P terms of A is q
$\therefore {{\text{S}}_{\text{p}}}=\text{q}$
Using the formula given in the hint for the sum we get
$\dfrac{\text{p}}{2}\ \left[ 2\text{a}+\left( \text{p}-1 \right)\text{d} \right]=\text{q}$ -(1)
Where a is first term of AP
D is the common difference
Similarly the sum of first q term of A is equal to q
$\therefore {{\text{S}}_{\text{q}}}=\text{p}$
$\dfrac{\text{q}}{2}\ \left[ 2\text{a}+\left( \text{q}-1 \right)\text{d} \right]=\text{p}$-(2)
Now subtracting equation (2) from equation (1), we get
\[\dfrac{\text{q}}{2}\ \left[ 2\text{a}+\left( \text{q}-1 \right)\text{d} \right]-\dfrac{\text{p}}{2}\ \left[ 2\text{a}+\left( \text{P}-1 \right)\text{d} \right]=\text{P}-\text{q}\]
\[\dfrac{\text{1}}{2}\ \left( 2\text{aq}+\text{q}\left( \text{q}-1 \right)\text{d}-2a\text{P}-\text{P}\left( \text{p}-1 \right)\text{d} \right)=\text{p}-\text{q}\]
Now taking the first and third tum together, we get
$\left( 2\text{aq}-2\text{ap} \right)+\text{d}\left( {{\text{q}}^{2}}-\text{q}-\left( {{\text{P}}^{2}}-\text{P} \right) \right)=2\left( \text{p}-\text{q} \right)$
$\Rightarrow 2a\left( \text{q}-\text{p} \right)+\text{d}\left( {{\text{q}}^{2}}-{{\text{P}}^{2}} \right)=2\left( \text{P}-\text{q} \right)$
Since we know that ${{\text{q}}^{\text{2}}}-{{\text{p}}^{\text{2}}}$ can be written as $\left( \text{q}-\text{P} \right)$ $\left( \text{q}+\text{P} \right)$
We can rewrite the equation as
$2a\left( \text{q}-\text{p} \right)+\text{d}\left[ \left( \text{q}-\text{p} \right)\left( \text{q}+\text{P} \right)-\left( \text{q}-\text{p} \right) \right]=2\left( \text{p}-\text{q} \right)$
$\therefore \left( \text{q}-\text{p} \right)$ is common in every term in left hand side, we can take it out
$\left( \text{q}-\text{p} \right)2\text{a}+\text{d}\left( \text{P}+\text{q}-\text{1} \right)=-2\left( \text{q}-\text{P} \right)$
Cancelling out $\left( \text{q}-\text{p} \right)$ on both side we get
2a+(p+q-1)d=-2-(3)
Now the sum of first $\left( \text{p}+\text{q} \right)$ terms can be written as
$S\text{p}+\text{q}=\dfrac{\text{P}+\text{q}}{2}\left[ 2\text{a}+\left( \text{P}+\text{q}-1 \right)\text{d} \right]$-(4)
Now substituting equation (3) in equation (4), we get
${{\text{S}}_{\text{p}+\text{q}}}=\dfrac{\text{p}+\text{q}}{2}\times -2$
${{\text{S}}_{\text{p}+\text{q}}}=-\left( \text{p}+\text{q} \right)$
Hence proved that the sum of first $\left( \text{p}+\text{q} \right)$ terms of an AP is equate to -$\left( \text{p}+\text{q} \right)$


Note: Alternatively you can find the solution by computing the value of a and d individually and then substituting it into the summation equation It is a simple approach but the calculation is complex.