Question

If $p^{th}$ terms is $q$ and $q^{th}$ term is $p$ of an arithmetic progression, then prove that ${(p+q)}^{th}$ term is zero.

Answer 1

$a_p=q$

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

$a+qd-d=q$ ----- (i)

$a_q=p$

$a+qd-d=p$ ----- (ii)

$a_{p+q}=a+\left[\right(p+q)-1]d$

$=a+[pd+qd-d]$ ---- (iii)

subtrct (ii) from (i), we get

$pd-qd=q-p$

$d=-1$

From (i)

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

$a=p+q-1$

put the values of $a$ and $d$ in (iii)

$a_{p+q}=p+q-1+(p+q-1)(-1)=p+q-1-p-q+1=0$

Hence proved