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Question: How many terms of the A.P. \(3,6,9,12,15...\)must be taken to make the sum \(108\)? A. \(6\) B. ...

How many terms of the A.P. 3,6,9,12,15...3,6,9,12,15...must be taken to make the sum 108108?
A. 66
B. 77
C. 88
D. 3636

Explanation

Solution

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 nn terms of an AP and then find out the value of nn for the given sum.

Complete step by step answer:
The given AP is 3,6,9,12,15...3,6,9,12,15.... The first term (a) of the A.P is 33. The common difference (d) of the A.P is 33
d=(63)=(96)=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 108108.
As we already know that the sum (Sn{S_n}) of an A.P is given by:
Sn=n2(2a+(n1)d){S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)
Here substituting the values of a=3a = 3 and d=3d = 3, we get,
108=n2(2×3+(n1)3)\Rightarrow 108 = \dfrac{n}{2}\left( {2 \times 3 + \left( {n - 1} \right)3} \right)
Now by simplifying the above equation we get,
108=n2(6+(n1)3) 108=n2(6+3n3) 108=n2(3n+3) 216=3n2+3n 3n2+3n216=0 \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 nn.
n=b±b24ac2an = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}...............(where a=3,b=3,c=216a = 3,b = 3,c = - 216)
n=3±324(3)(216)2(3) n=3±9+25926 n=3±26016 n=3±516 n=3+516,3516 n=486,546  \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 nn is not possible. So,
n=486=8\therefore n = \dfrac{{48}}{6} = 8
So, we are required to take 88 terms in an A.P to give a sum of 108108.

Hence, n=8n = 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.