Question

If the $p$th term of an A.P. is $q$ and the $q$th term is $p$, prove that the $n$th term is equal to $(p+q-n)$.

Answer 1

We know that the formula for the nth term is $t_n=a+(n-1)d$, where $a$ is the first term, $d$ is the common difference.

It is given that the $p$th term of an A.P is $q$ and $q$th term of an A.P is $p$, therefore,

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

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

Subtract equation 2 from 1 as follows:

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

Substitute the value of common difference $d$ in equation 1:

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

Now, the $n$th term with $a=q+p-1$ and $d=-1$ can be obtained as follows:

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

Hence, the $n$th term is $(p+q-n)$.