Question

If the sum of first p terms of an AP is the same as the sum of first q terms of an AP, then find the summation of first p+q terms.

Answer 1

Hint: In this question it is given that the sum of first p terms of an AP is the same as the sum of first q terms of an AP, then we have to find the summation of first p+q terms. So to find the solution we have to use the summation formula, i.e, the summation of first n terms,
 $$S_{n}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$.....(1)
Where, ‘a’ is the first term of AP and ‘d’ is the common difference.

Complete step-by-step solution:
Given that the sum of first p terms is equal to the sum of first q terms, so we can write,
 $$S_{p}=S_{q}$$
$$\Rightarrow \dfrac{p}{2} \left( 2a+\left( p-1\right) d\right) =\dfrac{q}{2} \left( 2a+\left( q-1\right) d\right) $$
$$\Rightarrow p\left( 2a+pd-d\right) =q\left( 2a+qd-d\right) $$
$$\Rightarrow 2ap+p^{2}d-pd=2aq+q^{2}d-qd$$
$$\Rightarrow 2ap-2aq+p^{2}d-q^{2}d-pd+qd=0$$
$$\Rightarrow 2a\left( p-q\right) +\left( p^{2}-q^{2}\right) d-\left( p-q\right) d=0$$
$$\Rightarrow 2a\left( p-q\right) +\left( p+q\right) \left( p-q\right) d-\left( p-q\right) d=0$$ [$$\because a^{2}-b^{2}=\left( a+b\right) \left( a-b\right) $$]
$$\Rightarrow 2a+\left( p+q\right) d-d=0$$ [ dividing both side by p+q]
$$\Rightarrow 2a+\left\{ \left( p+q\right) -1\right\} d=0$$......(2)

Now the first sum of (p+q) term,
$$S_{p+q}=\dfrac{p+q}{2} \left\{ 2a+\left( p+q-1\right) d\right\} $$
$$S_{p+q}=\dfrac{p+q}{2} \times 0$$ [ by equation (2)]
$$S_{p+q}=0$$
Therefore, the summation of first p+q terms is zero.

Note: While solving you have to know that you can also solve the above problem by using another formula, which states that, if you know the first and last term of an AP then the sum of first n terms $$S_{n}=\dfrac{n}{2} \left( a+l\right) $$
Where a is the first term of am AP and l is the last term.