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If the $p^{\text {th }}$ term of an A.P. be $q$ and $q^{\text {th }}$ term be $p$, then its $r^{\text {th }}$ term will be
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The correct answer is:
$p+q-r$
Given that, $\quad T_p=a+(p-1) d=q \ldots(1)$
and $T_q=a+(q-1) d=p \ldots(2)$
From (i) and (ii), we get $d=-\frac{(p-q)}{(p-q)}=-1$
Putting value of $d$ in equation (i), then $a=p+q-1$
Now, $r^{\text {th }}$ term is given by A.P.
$T_r=a+(r-1) d=(p+q-1)+(r-1)(-1)=p+q-r$
Note : Students should remember this question as a formula.
and $T_q=a+(q-1) d=p \ldots(2)$
From (i) and (ii), we get $d=-\frac{(p-q)}{(p-q)}=-1$
Putting value of $d$ in equation (i), then $a=p+q-1$
Now, $r^{\text {th }}$ term is given by A.P.
$T_r=a+(r-1) d=(p+q-1)+(r-1)(-1)=p+q-r$
Note : Students should remember this question as a formula.
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