According to the problem statement, the p^(th) term is a and the q^(th) term is b. We can write these as two equations using the formula for the n^(th) term.

The sum of the first n terms of an A.P. is given by S_n = (n)/(2)[2A + (n-1)D]. We need to find the sum of (p+q) terms, so we substitute n = p+q into this formula.

The $p^{\text {th }}$ term of an A.P. is $a$ and $q^{\text {th }}$ term is $b$ . Prove that the sum of its $(p+q)$ terms is $\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right] .$

Answer: The sum of its $(p+q)$ terms is $\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right]$ .

Step-by-step solution

Step 1: Define terms of an A.P.

Let $A$ be the first term and $D$ be the common difference of the Arithmetic Progression (A.P.). The $n^{\text{th}}$ term of an A.P. is given by the formula $t_n = A + (n-1)D$ .

Step 2: Formulate equations from given information

According to the problem statement, the $p^{\text{th}}$ term is $a$ and the $q^{\text{th}}$ term is $b$ . We can write these as two equations using the formula for the $n^{\text{th}}$ term.

Step 3: Solve for common difference D

Subtract equation (2) from equation (1) to eliminate $A$ . This allows us to solve for the common difference $D$ in terms of $a$ , $b$ , $p$ , and $q$ .

Step 4: Solve for the first term A

Substitute the value of $D$ back into equation (1) to find the first term $A$ . This expresses $A$ in terms of $a$ , $b$ , $p$ , and $q$ .

Step 5: Apply sum formula for (p+q) terms

The sum of the first $n$ terms of an A.P. is given by $S_n = \frac{n}{2}[2A + (n-1)D]$ . We need to find the sum of $(p+q)$ terms, so we substitute $n = p+q$ into this formula.

Step 6: Substitute A and D and simplify

Substitute the expressions for $A$ and $D$ into the sum formula. Then, simplify the terms inside the square brackets. Combine the terms involving $\frac{a-b}{p-q}$ and rearrange to match the required form.

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The pth p^{\text {th }}pth term of an A.P. is aaa and qth q^{\text {th }}qth term is bbb. Prove that the sum of its (p+q)(p+q)(p+q) terms is p+q2[a+b+a−bp−q].\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right] .2p+q​[a+b+p−qa−b​].