In an AP, the pth term is q and (p + q)th term is 0.


In an AP, the $p^{\text {th }}$ term is $q$ and $(p+q)^{\text {th }}$ term is 0 . Show that its $q^{\text {th }}$ term is $p$.


Given: $p^{\text {th }}$ term is $q$ and $(p+q)^{\text {th }}$ term is 0 .

To prove: $q^{\text {th }}$ term is $p$.

$p^{\text {th }}$ term is given by

$q=a+(p-1) \times d \ldots \ldots$ equation 1

$(p+q)^{\text {th }}$ term is given by

$0=a+(p+q-1) \times d$

$0=a+(p-1) \times d+q \times d$

Using equation1

$0=q+q \times d$


Put in equation1 we get


$q^{\text {th }}$ term is

$\Rightarrow q+p-1+(q-1) \times(-1)$

$\Rightarrow p$

Hence proved.

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