Question

How many terms of the A.P. \[ - 6\], \[ - \dfrac{{11}}{2}\], \[ - 5\], … are needed to give the sum \[ - 25\]?

Answer 1

Hint: First we will find the common difference and then use the formula of sum of \[n\] terms of the arithmetic progression A.P., that is, \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference. Apply this formula, and then substitute the value of \[a\] and \[d\] in the obtained equation. Then we will factorize the equation to find the value of \[n\].

Complete step by step solution:
We are given that the A.P. \[ - 6\], \[ - \dfrac{{11}}{2}\], \[ - 5\], … and its sum is \[ - 25\].
We know that the arithmetic progression is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.

We will now find the first term \[a\] and the second term \[b\] of the given A.P.

\[a = - 6\]
\[b = - \dfrac{{11}}{2}\]

Subtracting the second term from the first term to find the common difference \[d\] of the given A.P., we get

\[
   \Rightarrow d = - \dfrac{{11}}{2} - \left( { - 6} \right) \\
   \Rightarrow d = - \dfrac{{11}}{2} + 6 \\
   \Rightarrow d = \dfrac{{ - 11 + 12}}{2} \\
   \Rightarrow d = \dfrac{1}{2} \\
 \]

We will use the formula of sum of \[n\] terms of the arithmetic progression A.P., that is, \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference.

We know that \[{S_n} = - 25\].

Substituting these values of \[{S_n}\], \[a\] and \[d\] in the above formula for the sum of the arithmetic progression, we get

\[
   \Rightarrow - 25 = \dfrac{n}{2}\left( {2\left( { - 6} \right) + \left( {n - 1} \right)\dfrac{1}{2}} \right) \\
   \Rightarrow - 25 = \dfrac{n}{2}\left( { - 12 + \dfrac{{n - 1}}{2}} \right) \\
   \Rightarrow - 25 = \dfrac{n}{2}\left( {\dfrac{{ - 24 + n - 1}}{2}} \right) \\
   \Rightarrow - 25 = \dfrac{n}{2}\left( {\dfrac{{ - 25 + n}}{2}} \right) \\
   \Rightarrow - 25 = \dfrac{n}{4}\left( {n - 25} \right) \\
   \Rightarrow - 25 = \dfrac{{{n^2} - 25n}}{4} \\
 \]

Multiplying the above equation by 4 on each of the sides, we get

\[
   \Rightarrow - 25\left( 4 \right) = 4\left( {\dfrac{{{n^2} - 25n}}{4}} \right) \\
   \Rightarrow - 100 = {n^2} - 25n \\
 \]

Adding the above equation by 100 on each of the sides, we get

\[
   \Rightarrow - 100 + 100 = {n^2} - 25n + 100 \\
   \Rightarrow 0 = {n^2} - 25n + 100 \\
   \Rightarrow {n^2} - 25n + 100 = 0 \\
 \]

Factorizing the above equation to find the value of \[n\], we get

\[
   \Rightarrow {n^2} - 20n - 5n + 100 = 0 \\
   \Rightarrow n\left( {n - 20} \right) - 5\left( {n - 20} \right) = 0 \\
   \Rightarrow \left( {n - 20} \right)\left( {n - 5} \right) = 0 \\
 \]
\[ \Rightarrow n - 20 = 0\] or \[n - 5 = 0\]
\[ \Rightarrow n = 20\] or \[n = 5\]

Thus, the total number of terms needed is 20 or 5.

Note:
In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and factorization of quadratic equations. Some students use the formula to find the sum, \[S = \dfrac{n}{2}\left( {a + l} \right)\], where \[l\] is the last term, but we do not have the value of \[l\]. So, we will here use \[S = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference.