Question

If the $p^{th}$ term of an A.P. be q and $q^{th}$ term be p, then its $r^{th}$ term will be

• A. p + q + r
• B. p - q + r
• C. p + q - r
• D. p - q - r

Answer 1

Answer: (C)

p+q-r

Answer 2

Let the first term of the A.P. be 'a' and the common difference be 'd'. The general formula for the $n^{th}$ term of an A.P. is given by: $t_n = a + (n-1)d$

According to the given conditions:

1. The $p^{th}$ term is $q$: $t_p = a + (p-1)d = q$ --- (Equation 1)
2. The $q^{th}$ term is $p$: $t_q = a + (q-1)d = p$ --- (Equation 2)

Subtract Equation 2 from Equation 1: $(a + (p-1)d) - (a + (q-1)d) = q - p$
$a + pd - d - a - qd + d = q - p$
$pd - qd = q - p$
$d(p - q) = -(p - q)$

Assuming $p \neq q$, we can divide both sides by $(p-q)$:
$d = \frac{-(p-q)}{(p-q)}$
$d = -1$

Now substitute the value of $d = -1$ into Equation 1:
$a + (p-1)(-1) = q$
$a - (p-1) = q$
$a - p + 1 = q$
$a = p + q - 1$

So, the first term of the A.P. is $a = p + q - 1$ and the common difference is $d = -1$.

Finally, we need to find the $r^{th}$ term, $t_r$. Using the general formula for the $n^{th}$ term:
$t_r = a + (r-1)d$

Substitute the values of 'a' and 'd' we found:
$t_r = (p + q - 1) + (r-1)(-1)$
$t_r = p + q - 1 - r + 1$
$t_r = p + q - r$

Thus, the $r^{th}$ term of the A.P. is $p+q-r$.