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If the sum of first p terms of an A.P is equal to the sum of first q terms, then show that the sum of its first (p + q) terms is zero where $p \ne q$.

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Hint: Use the direct formula to find the sum of first p terms of an A.P, equate it with the sum of first q terms of this A.P. Now the first term and the common difference will remain the same in both the cases as we are talking about only a single A.P series, however the sum of the number of terms is only varying.

Complete step-by-step answer:
Now, if we have an A.P then the sum of its first n terms is given as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where $a$ is the first term, d is the common difference and n is the total number of terms……………………. (1)

Thus, the sum of first p terms of an A.P using equation (1) can be written as
${S_p} = \dfrac{p}{2}\left( {2a + \left( {p - 1} \right)d} \right)$…………………… (2)

Now, the sum of first q terms of this same A.P using equation (1) can be written as
${S_q} = \dfrac{q}{2}\left( {2a + \left( {q - 1} \right)d} \right)$………………… (3) As we are talking about the same A.P hence the value of first term and common difference remained the same in both equation (2) and equation (3).

Now, in question it is given that sum of first p terms of an A.P is equal to the sum of first q terms hence equation (2) = equation (3)
$ \Rightarrow {S_p} = {S_q}$

Hence, substituting the values we get
$\dfrac{p}{2}\left( {2a + \left( {p - 1} \right)d} \right) = \dfrac{q}{2}\left( {2a + \left( {q - 1} \right)d} \right)$

On simplifying we get,
$p\left( {2a + \left( {p - 1} \right)d} \right) = q\left( {2a + \left( {q - 1} \right)d} \right)$
$ \Rightarrow 2ap + {p^2}d - pd = 2aq + {q^2}d - qd$

Now let’s take similar terms to same side and take common we get
$ \Rightarrow 2a\left( {p - q} \right) + d\left( {{p^2} - {q^2}} \right) - d\left( {p - q} \right) = 0$
Using ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ we get
$ \Rightarrow 2a\left( {p - q} \right) + \left( {p + q} \right)\left( {p - q} \right)d - d\left( {p - q} \right) = 0$

Let’s take p-q common
$ \Rightarrow \left( {p - q} \right)\left[ {2a + \left( {p + q} \right)d - d} \right] = 0$
Now this equation gets satisfied if p – q = 0 or the second term that is $\left[ {2a + \left( {p + q} \right)d - d} \right]$ is equal to zero.
However in the question it is stated that $p \ne q$ thus p-q=0 doesn’t satisfy thus we are left with only one equation that is $\left[ {2a + \left( {p + q} \right)d - d} \right] = 0$…………………….. (4)

Now we have to show that the sum of first (p + q) terms is zero.
So Sum of first (p + q) terms is equal to
$ \Rightarrow {S_{p + q}}$
Now using equation (1) ${S_{p + q}} = \dfrac{{p + q}}{2}\left( {2a + \left( {p + q - 1} \right)d} \right)$
This can be written as
$ \Rightarrow {S_{p + q}} = \dfrac{{p + q}}{2}\left( {2a + \left( {p + q} \right)d - d} \right)$
Using equation (4) we know that $\left[ {2a + \left( {p + q} \right)d - d} \right] = 0$

Hence ${S_{p + q}} = 0$

Note: Whenever we face such type of problems the key point here is to simplify use the basic sum of n terms of an arithmetic progression, we need to take care of the fact that if we are talking about sum up to different numbers however of the same A.P, then obviously the first term and the common difference won’t change.