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  • 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).

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).

Answers (1)

The sum of n terms is given by

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

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

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


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


  \\ S_{p+q}=\frac{p+q}{2} \left( 2a+ \left( p+q-1 \right) d \right) \\\\ =\frac{p}{2} \left( 2a+ \left( p-1 \right) d+qd \right) +\frac{q}{2} \left( 2a+ \left( q-1 \right) d+pd \right) \\\\
 \\ =\frac{p}{2} \left( 2a+ \left( p-1 \right) d \right) +\frac{pqd}{2}+\frac{q}{2} \left( 2a+ \left( q-1 \right) d \right) +\frac{pqd}{2} \\\\ =q+p+pqd \\\\


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


\\ =\frac{p}{2} \left( 2a+ \left( p-1 \right) d \right) -\frac{pqd}{2}-\frac{q}{2} \left( 2a+ \left( p-1 \right) d \right) +\frac{q^{2}d}{2} \\\\ =q-\frac{pqd}{2}- \left( \frac{q}{p} \right) \left( \frac{p}{2} \right) \left( 2a+ \left( p-1 \right) d \right) +\frac{q^{2}d}{2} \\\\ =q-\frac{q^{2}}{p}-\frac{pqd}{2}+\frac{q^{2}d}{2} \\\\ =\frac{ \left( pq-q^{2} \right) }{p}+\frac{d \left( q^{2}-pq \right) }{2} \\\\ =\frac{pq-q^{2}}{p}-\frac{pq-q^{2}}{2} \left( -\frac{2 \left( p+q \right) }{pq} \right) \\\\  
\\ =\frac{pq-q^{2}}{p}+ \left( pq-q^{2} \right) \left[ \frac{1}{q}+\frac{1}{p} \right] \\\\ =\frac{2 \left( pq-q^{2} \right) }{p}+\frac{pq-q^{2}}{q} \\\\ =\frac{2q \left( p-q \right) }{p}+p-q \\\\

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