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If $p$ times the $p$ th term of an AP is $q$ times the $q$ th term, then what is the $(p+q)$ th term equal to? $[2010-I]$
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Let a and d be the first term and common difference of an AP respectively $\mathrm{p}^{\text {th }}$ term $=\mathrm{a}+(\mathrm{p}-1) \mathrm{d}$
and $q^{\text {th }}$ term $=\mathrm{a}+(\mathrm{q}-1) \mathrm{d}$
According to question.
$p[a+(p-1) d]=q[a+(q-1) d]$
$\Rightarrow p a+\left(p^{2}-p\right) d=q a+\left(q^{2}-q\right) d$
$\Rightarrow(p-q) a=\left(q^{2}-p^{2}+p-q\right) d$
$\Rightarrow(p-q) a=(p-q)(-p-q+1) d$
$\Rightarrow a=-(p+q-1) d$
Now, $(p+q)^{t h}$ term $=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) d$
$\quad=-(p+q-1) d+(p+q-1) d=0$
and $q^{\text {th }}$ term $=\mathrm{a}+(\mathrm{q}-1) \mathrm{d}$
According to question.
$p[a+(p-1) d]=q[a+(q-1) d]$
$\Rightarrow p a+\left(p^{2}-p\right) d=q a+\left(q^{2}-q\right) d$
$\Rightarrow(p-q) a=\left(q^{2}-p^{2}+p-q\right) d$
$\Rightarrow(p-q) a=(p-q)(-p-q+1) d$
$\Rightarrow a=-(p+q-1) d$
Now, $(p+q)^{t h}$ term $=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) d$
$\quad=-(p+q-1) d+(p+q-1) d=0$
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