
How many terms of the A.P. $3,6,9,12,15...$must be taken to make the sum $108$?
A. $6$
B. $7$
C. $8$
D. $36$
Answer
548.7k+ views
Hint:The numbers of a series are said to be in A.P if and only if the common difference (that is, the difference between the consecutive terms) remains constant throughout the series. Next, use the direct formula for the sum of $n$ terms of an AP and then find out the value of $n$ for the given sum.
Complete step by step answer:
The given AP is $3,6,9,12,15...$. The first term (a) of the A.P is $3$. The common difference (d) of the A.P is $3$
$ \Rightarrow d = \left( {6 - 3} \right) = \left( {9 - 6} \right) = 3$
We have to find out the number of terms in the A.P whose sum equals to be $108$.
As we already know that the sum (${S_n}$) of an A.P is given by:
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Here substituting the values of $a = 3$ and $d = 3$, we get,
$ \Rightarrow 108 = \dfrac{n}{2}\left( {2 \times 3 + \left( {n - 1} \right)3} \right)$
Now by simplifying the above equation we get,
$\Rightarrow 108 = \dfrac{n}{2}\left( {6 + \left( {n - 1} \right)3} \right) \\
\Rightarrow 108 = \dfrac{n}{2}\left( {6 + 3n - 3} \right) \\
\Rightarrow 108 = \dfrac{n}{2}\left( {3n + 3} \right) \\
\Rightarrow 216 = 3{n^2} + 3n \\
\Rightarrow 3{n^2} + 3n - 216 = 0 \\
$
Now, we have a quadratic equation so we apply the quadratic formula to find out the value of $n$.
$n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$...............(where $a = 3,b = 3,c = - 216$)
$
\Rightarrow n = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 3 \right)\left( { - 216} \right)} }}{{2\left( 3 \right)}} \\
\Rightarrow n = \dfrac{{ - 3 \pm \sqrt {9 + 2592} }}{6} \\
\Rightarrow n = \dfrac{{ - 3 \pm \sqrt {2601} }}{6} \\
\Rightarrow n = \dfrac{{ - 3 \pm 51}}{6} \\
\Rightarrow n = \dfrac{{ - 3 + 51}}{6},\dfrac{{ - 3 - 51}}{6} \\
\Rightarrow n = \dfrac{{48}}{6},\dfrac{{ - 54}}{6} \\
$
As we know that a negative value of $n$ is not possible. So,
$\therefore n = \dfrac{{48}}{6} = 8$
So, we are required to take $8$ terms in an A.P to give a sum of $108$.
Hence, $n = 8$ is the required answer and option C is correct.
Note:Whenever faced with such types of questions, it is advised to have good knowledge of general formulas related to the series which is asked in the question. Here in order to solve this question, we must have knowledge of arithmetic progression. This will not only help save a lot of time but also to get the right answer.
Complete step by step answer:
The given AP is $3,6,9,12,15...$. The first term (a) of the A.P is $3$. The common difference (d) of the A.P is $3$
$ \Rightarrow d = \left( {6 - 3} \right) = \left( {9 - 6} \right) = 3$
We have to find out the number of terms in the A.P whose sum equals to be $108$.
As we already know that the sum (${S_n}$) of an A.P is given by:
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Here substituting the values of $a = 3$ and $d = 3$, we get,
$ \Rightarrow 108 = \dfrac{n}{2}\left( {2 \times 3 + \left( {n - 1} \right)3} \right)$
Now by simplifying the above equation we get,
$\Rightarrow 108 = \dfrac{n}{2}\left( {6 + \left( {n - 1} \right)3} \right) \\
\Rightarrow 108 = \dfrac{n}{2}\left( {6 + 3n - 3} \right) \\
\Rightarrow 108 = \dfrac{n}{2}\left( {3n + 3} \right) \\
\Rightarrow 216 = 3{n^2} + 3n \\
\Rightarrow 3{n^2} + 3n - 216 = 0 \\
$
Now, we have a quadratic equation so we apply the quadratic formula to find out the value of $n$.
$n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$...............(where $a = 3,b = 3,c = - 216$)
$
\Rightarrow n = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 3 \right)\left( { - 216} \right)} }}{{2\left( 3 \right)}} \\
\Rightarrow n = \dfrac{{ - 3 \pm \sqrt {9 + 2592} }}{6} \\
\Rightarrow n = \dfrac{{ - 3 \pm \sqrt {2601} }}{6} \\
\Rightarrow n = \dfrac{{ - 3 \pm 51}}{6} \\
\Rightarrow n = \dfrac{{ - 3 + 51}}{6},\dfrac{{ - 3 - 51}}{6} \\
\Rightarrow n = \dfrac{{48}}{6},\dfrac{{ - 54}}{6} \\
$
As we know that a negative value of $n$ is not possible. So,
$\therefore n = \dfrac{{48}}{6} = 8$
So, we are required to take $8$ terms in an A.P to give a sum of $108$.
Hence, $n = 8$ is the required answer and option C is correct.
Note:Whenever faced with such types of questions, it is advised to have good knowledge of general formulas related to the series which is asked in the question. Here in order to solve this question, we must have knowledge of arithmetic progression. This will not only help save a lot of time but also to get the right answer.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

