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If the sum of p terms of an A.P. is q and

Question:

 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:

The sum of $n$ terms of an $A P$ is given by

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

Where $a$ is the first term and $d$ is the common difference Given that $S_{p}=q$ and $S_{q}=p$

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

We know $S_{p}=q$

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

On rearranging we get

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

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

Again we have $S_{q}=p$

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

On rearranging we get

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

Subtract (i) from (ii) that is (ii) - (i)

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

Subtract (i) from (ii) that is (ii) - (i)

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

On simplifying we get

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

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

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

Computing and simplifying we get

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

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

We have to show that $S_{p+q}=-(p+q)$

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

Above equation can be written as

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

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

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

Using $(\mathrm{m})$ and $(\mathrm{n})$

$\Rightarrow S_{p+q}=S_{p}+S_{q}+p q d$

$=q+p+p q d$

Substitute d from (iii)

$\Rightarrow S_{p+q}=q+p+p q\left(-2 \frac{(p+q)}{p q}\right)$

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

$=-(p+q)$

Now we have to find sum of $p-q$ terms that is $S_{p-q}$

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

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

The above equation can be written as

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

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

The above equation can be written as

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

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

Using $(\mathrm{m})$ and $(\mathrm{n})$

$=\mathrm{S}_{\mathrm{p}}-\frac{\mathrm{pqd}}{2}-\frac{\mathrm{q}}{2} \frac{2 \mathrm{~S}_{\mathrm{p}}}{\mathrm{p}}+\frac{\mathrm{q}^{2} \mathrm{~d}}{2}$

Substituting the value of $S_{p}=q$ we get

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

$=q-\frac{q^{2}}{p}+\left(\frac{q^{2}-q p}{2}\right) d$

Substitute d from (iii)

$=q-\frac{q^{2}}{p}+\left(\frac{q^{2}-q p}{2}\right)\left(-2 \frac{(p+q)}{p q}\right)$

Simplifying and computing we get

$=\frac{q p-q^{2}}{p}-\left(q^{2}-q p\right)\left(\frac{p+q}{p q}\right)$

$=\frac{q p-q^{2}}{p}+\left(q p-q^{2}\right)\left(\frac{p+q}{p q}\right)$

$=\frac{q p-q^{2}}{p}+\left(q p-q^{2}\right)\left(\frac{1}{p}+\frac{1}{q}\right)$

$=\frac{q p-q^{2}}{p}+\frac{\left(q p-q^{2}\right)}{p}+\frac{\left(q p-q^{2}\right)}{q}$

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

Hence sum of $p-q$ terms is $2 \frac{q(p-q)}{p}+p-q$

 

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