We are given that the p^(th) term is q and the q^(th) term is p. Substituting n = p and n = q into the n^(th) term formula gives us a system of two linear equations in a and d.

if the teeth term of an ap is Q and the cute term is P then its P + Q to term is

Answer: 0

Step-by-step solution

Step 1: State the formula for the nth term of an AP

Let $a$ be the first term and $d$ be the common difference of the given AP. The general formula for the $n^{\text{th}}$ term is $a_n = a + (n-1)d$ .

Step 2: Set up equations for the p-th and q-th terms

We are given that the $p^{\text{th}}$ term is $q$ and the $q^{\text{th}}$ term is $p$ . Substituting $n = p$ and $n = q$ into the $n^{\text{th}}$ term formula gives us a system of two linear equations in $a$ and $d$ .

Step 3: Subtract the equations to solve for d

Subtracting equation (2) from equation (1) eliminates the first term $a$ . Factoring out $d$ yields $(p-q)d = q - p = -(p-q)$ . Since $p \neq q$ , dividing both sides by $(p-q)$ gives $d = -1$ .

Step 4: Find the value of the first term a

Substitute $d = -1$ back into equation (1). Expanding the expression gives $a - p + 1 = q$ , which leads directly to $a = p + q - 1$ .

Step 5: Determine the (p+q)-th term

We use the $n^{\text{th}}$ term formula for $n = p+q$ , which is $a_{p+q} = a + (p+q-1)d$ . Substituting $a = p+q-1$ and $d = -1$ gives $(p+q-1) - (p+q-1) = 0$ .

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