Question

plzz do fast. ........
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Answer 1

$p^{th}$ term is $q$ and $q^{th}$ term is $p$

$n^{th}$ term of an AP is

$a_n = a+(n-1)d$

where a is the first term and d is the common difference

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

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

Finding the difference of the two equations, we get

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

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

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

$d = (-1)$

Substituting the value of d in any of the two equations

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

$q=a-p+1$

$a=p+q-1$

Substituting the values of a and d in the equation for $n^{th}$ term of an AP, we get

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

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

$a_n=p+q-n$

Option (A) is the answer

HOPE IT HELPS!!