Question

How many terms of AP: 27, 24, 21, ……… should be taken so that their sum is zero? What is the value of that last term?

Answer 1

Hint: In this question, we will use the concept of arithmetic progression (A.P). we will use the formula of finding the sum of all the terms of an A.P. where first term is ‘a’ and the common difference is ‘d’ which is given by
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$.

Complete step by step answer:
A sequence \[{a_1},{a_2},{a_3},.........,{a_n}\] is called an arithmetic progression if the first term is a and the common difference is d, then a, a+d, a+2d, a+3d, ………. is an arithmetic progression.
We know that the sum of all terms of an A.P. is given by,
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$, ………(i)
this can also be written as:
${S_n} = \dfrac{n}{2}[a + l]$, ……….(ii)
here l = last term = a + (n-1)d.
Here given that, a = 27 and ${S_n} = 0$
We know that d = common difference = ${a_2} - {a_1} = 24 - 27 = - 3$.
Putting these values in equation (i), we get
\[
   \Rightarrow 0 = \dfrac{n}{2}[2 \times 27 + (n - 1)( - 3)] \\
   \Rightarrow 0 = \dfrac{n}{2}[54 - 3n + 3] \\
   \Rightarrow 0 = \dfrac{n}{2}[57 - 3n] \\
   \Rightarrow 0 = n[57 - 3n] \\
   \Rightarrow 3{n^2} - 57n = 0 \\
 \]
Taking 3n common, we get
N - 19 = 0
n = 19 or 0 because we took 3n common so from where we will get 0.
n = 0 cannot be possible, so n = 19.
Now, we have to find the last term.
Put all the given values in equation (ii), we get
$
   \Rightarrow {S_n} = \dfrac{n}{2}[a + l] \\
   \Rightarrow 0 = \dfrac{{19}}{2}[27 + l] \\
   \Rightarrow l = - 27 \\
$

Hence, we can say that the number of terms in the given A.P. is 19 and the value of the $19^{th}$ term or last term is -27.

Note: Whenever we ask such types of questions, we will use the basic formulae related to the arithmetic progression. First we have to find out the details given in the question and then by using them we will find out the value of d. after that we will find the number of terms by using the formula of sum of all terms and then we can also find the last term. Through this we will get the required answer.