If the sum of p terms of an A.P. is q and the sum of q terms is p, show that the sum of p + q terms is – (p + q). Also, find the sum of first p

If the sum of p terms of an A.P. is q and the sum of q terms is p, show that the sum of p + q terms is – (p + q). Also, find the sum of first p – q terms (p > q).

CBSE | Class 11 | Maths | NCERT Exemplar | Sequences and Series

If the sum of p terms of an A.P. is q and the sum of q terms is p, show that the sum of p + q terms is – (p + q). Also, find the sum of first p – q terms (p > q).

Solution:

An AP’s sum of n terms is given by

${S_{n}}={\frac{n}{2}}(2a+(n-1)d)$

Where the first term is ‘a’ and the common difference is ‘d’.

It is given that $S_{p}=q$ and $S_{q}=p$

$\Rightarrow \mathrm{S}_{\mathrm{p}}=\frac{\mathrm{p}}{2}(2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d})_{\ldots 1}$

It is known that $S_{p}=q$

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

Now rearrange

$\Rightarrow \frac{2 q}{p}=2 a+(p-1) d \ldots(i)$

$\Rightarrow S_{q}=\frac{q}{2}(2 a+(q-1) d) \ldots \ldots .2$

Again we have $S_{q}=p$

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

When we rearrange we obtain

$\Rightarrow \frac{2 p}{q}=2 a+(q-1) d \ldots(i i)$

Subtract eq.(i) from eq.(ii) i.e. eq.(ii) – eq.(i)

$\Rightarrow \frac{2 p}{q}-\frac{2 q}{p}=(q-1) d-(p-1) d$

Subtract eq.(i) from eq.(ii) i.e. eq.(ii) – eq.(i)

$\Rightarrow \frac{2 p}{q}-\frac{2 q}{p}=(q-1) d-(p-1) d$

When we simplify we obtain

$\Rightarrow 2 \frac{p^{2}-q^{2}}{p q}=(q-1-p+1) d$

Using the $a^{2}-b^{2}=(a+b)(a-b)$ formula we obtain

$\Rightarrow 2 \frac{(p+q)(p-q)}{p q}=(q-p) d$

Compute and simplify now,

$\Rightarrow 2 \frac{-(p+q)(q-p)}{p q}=(q-p) d$

$\Rightarrow-2 \frac{(p+q)}{p q}=d \ldots(i i i)$

We need to show that $\mathrm{S}_{\mathrm{p}+\mathrm{q}}=-(\mathrm{p}+\mathrm{q})$

$\mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}(2 a+(\mathrm{p}+\mathrm{q}-1) \mathrm{d})$

Above eq. can also be re-written as

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

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

$=\frac{\mathrm{p}}{2}(2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d})+\frac{\mathrm{pqd}}{2}+\frac{\mathrm{q}}{2}(2 \mathrm{a}+(\mathrm{q}-1) \mathrm{d})+\frac{\mathrm{qpd}}{2}$

Now using the $(\mathrm{m}) \text { and }(\mathrm{n})$

$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\mathrm{S}_{p}+\mathrm{S}_{\mathrm{q}}+\mathrm{p} \mathrm{q} \mathrm{d}$

$=\mathrm{q}+\mathrm{p}+\mathrm{p} \mathrm{q} \mathrm{d}$

From eq.(iii) substitute ‘d’ $\Rightarrow S_{p+q}=q+p+p q\left(-2 \frac{(p+q)}{p q}\right)$

$=(p+q)-2(p+q)$

$=-(p+q)$

We now need to find the sum of $p-q$ terms that is $S_{p-q}$

$\Rightarrow \mathrm{S}_{\mathrm{p}-\mathrm{q}}=\frac{\mathrm{p}-\mathrm{q}}{2}(2 \mathrm{a}+(\mathrm{p}-\mathrm{q}-1) \mathrm{d})$

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

The above eq. can also be re-written as

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

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

The above eq. can also be re-written as

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

$=\frac{\mathrm{p}}{2}(2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d})-\frac{\mathrm{pqd}}{2}-\frac{\mathrm{q}}{2}(2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d})+\frac{\mathrm{q}^{2} \mathrm{~d}}{2}$