Question

In an AP, of which a is the first term, if the sum of the first p terms is zero, show that the sum of the next q terms is $ - \dfrac{{a(p + q)q}}{{p - 1}}$ .

Answer 1

Hint: In this question, we will first write the formula to find the sum of n terms of an AP and later will find the sum of the first p terms and then equate it to zero to get a relation. Next we will find the (p+1)th term and then find the sum of next q terms.

Complete step-by-step answer:
In the question, we have an AP whose first term = a .
Let the common difference be ‘d’.
We know that sum of ‘n’ terms of an AP is given by,
$S = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$ .
Therefore, sum of first p terms is calculated as,
$S = \dfrac{p}{2}\left( {2a + (p - 1)d} \right)$ .
On equation this , sum to zero, we get,
$\dfrac{p}{2}\left( {2a + (p - 1)d} \right)S = 0$
On further solving, we get,
$\left( {2a + (p - 1)d} \right) = 0$
$ \Rightarrow d = \dfrac{{ - 2a}}{{p - 1}}$ (1)
Now, let us find the (p+1)th term.
WE know that nth term of an AP is given by,
${T_n} = a + (n - 1)d$ .
Putting the value in the above equation. We get,
${T_{p + 1}} = a + (p + 1 - 1)d = a + pd$. This is the first term for next q terms.
Again using the formula for finding the sum of next q terms, we get,
$S = \dfrac{q}{2}\left( {2(a + pd) + (q - 1)d} \right)$
On rearranging the terms, we get,
$S = \dfrac{q}{2}\left( {2a + (p - 1)d + (p + q)d} \right)$
Putting the value of d from equation 1 we get,
$S = \dfrac{q}{2}\left( {0 - (p + q)\dfrac{{2a}}{{p - 1}}} \right) = - a\dfrac{{(p + q)q}}{{p - 1}}$

Note- In this question, you should remember the formula to find the nth term of an AP and also the formula to find the sum of nth term of the AP. The arithmetic mean of first n terms in an AP is equal to sum of terms divided by number of terms.