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How many terms of the series $9,12,15,.....$ must be taken to make the sum of terms $360?$

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Hint: Identify the series and apply the formula for the sum of first $n$ terms of the series.
The given series is $9,12,15,.....$
Difference between any two consecutive terms of the series is constant. Thus, the series is in A.P.
First term, $a = 9,$Common Difference, $d = 3.$
Now, we need the sum of terms to be up to $360.$
We know that the sum of first $n$terms of A.P. is${S_n} = \frac{n}{2}[2a + (n - 1)d]$. Using this, we’ll get:
\[
   \Rightarrow 360 = \frac{n}{2}[2(9) + (n - 1) \times 3] \\
   \Rightarrow n(3n + 15) = 720 \\
   \Rightarrow n(n + 5) = 240 \\
   \Rightarrow {n^2} + 5n - 240 = 0 \\
\]
We know that the solution of a quadratic equation is$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Using this for the above equation, we’ll get:
$
   \Rightarrow n = \frac{{ - 5 \pm \sqrt {25 - 4(1)( - 240)} }}{{2(1)}} \\
   \Rightarrow n = \frac{{ - 5 \pm \sqrt {985} }}{2} \\
$
But the number of terms cannot be negative. So, we will ignore the negative value of $n$.
Then we have:
$
   \Rightarrow n = \frac{{ - 5 + \sqrt {985} }}{2} \\
   \Rightarrow n = 13.19 \\
   \Rightarrow n \approx 14. \\
$
Therefore, for having the sum of the series at least 360, the series must contain a minimum of 14 terms.
Note: We can also determine the sum of the series using ${S_n} = \frac{n}{2}(a + l)$, where $l$ is the last term of the series under consideration. So, for using this formula, we have to find out the last term of the series.