Question

If the $pth$ term of an $AP$ is $q$ and the $qth$ term $p$, then prove that its $nth$ term is $(p+1-n)$ and hence prove that its $(p+q{)}^{th}$ term is zero.

Answer 1

We know, $n^{th}$ term $a_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$a_n=n^{th}$ term

Given,

$P^{th}$ term is q

$q=a+(p-1)d...\left(1\right)$

$q^{th}$ term is p

$p=a+(q-1)d...\left(2\right)$

(1) - (2)

$q-p=(p-1-q+1)d$

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

put $d=-1$ in (1)

$q=a+(p-1)(-1)$

$a=q+p-1$

$n^{th}$ term $=a+(n-1)d$

$=q+p-1+(n-1)(-1)$

$=q+p-1-n+1$

$=p+q-n$

$(p+q{)}^{th}$ term $=a+(p+q-1)d$

$=p+q-1+(p+q-1)(-1)$

$=0$