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If $p^{th},\:q^{th}\:\:and\:\:r^{th}$ terms oa an A.P. are $a,\:b\:and\:c$ respectively, then show that $(q-r)a+(r-p)b+(p-q)c=0$

1 Answer

Given $p^{th}$ term of an A.P.$a$, $q^{th}$ term $=b$ and $r^{th}$ term $=c$
$\Rightarrow\:t_p=A+(p-1)D=a$ where $A$ is first term and $D$ is common difference
$t_q=A+(q-1)D=b$ and $A+(r-1)D=c$
Step 2
Step 3
Similarly $(p-q)c=(p-q).\big[A+(r-1)D\big]$
Step 4
Adding the three we get $(q-r)a+(r-p)b+(p-q)c=$
Hence proved.
answered Mar 31, 2014 by rvidyagovindarajan_1

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