Question

If the pth term of an AP is q and the qth term is p, prove that its nth term is $(p + q - n)$.

Answer 1

The formula of nth term of AP is as follow:
${a_n} = a + (n - 1)d$
where ‘an’ represents nth term of AP,
‘a’ represents first term of AP,
‘n’ represents number of terms and
‘d’ represents common difference

Complete step-by-step answer:
The nth term of AP is written as ${a_n} = a + (n - 1)d$
It is given that pth term of an AP is q which means
${a_p} = a + (p - 1)d = q$ --1
And also it is given that qth term of AP is p which is written as
${a_q} = a + (q - 1)d = p$ --2
Now we got 2 linear equations in terms of ‘a’ and ‘d’
Subtract equation 2 from equation 1; we get
$\Rightarrow$ $(p - 1)d - (q - 1)d = q - p$
Taking ‘d’ as common, we get
$\Rightarrow$ $(p - 1 - q + 1)d = q - p$
$\Rightarrow$ $(p - q)d = q - p$
If we take (-1) common from RHS, we get
$\Rightarrow$ $(p - q)d = - (p - q)$
Which gives $d = - 1$
Put this value of ‘d’ in equation 1; we get
$\Rightarrow$ $a + (p - 1)( - 1) = q$
Which simplifies to $a = q + p - 1$
Now, we get values of both ‘a’ and ‘d’
Now, we can find value of nth term
$\Rightarrow$ ${a_n} = a + (n - 1)d$
Put values of ‘a’ and ‘d’; we get
$\Rightarrow$ ${a_n} = q + p - 1 + (n - 1)( - 1)$
Which simplifies to ${a_n} = q + p - n$

Note: To find the value of ‘n’ number of variables, we must have ‘n’ number of equations.
It means if we have two variables ‘x’ and ‘y’ then we must have two equations in ‘x’ and ‘y’ to calculate the value of them. So, to find any AP, we need first term and common difference, so we only need two equations in ‘a’ and ‘d’ to form an AP.