# How many terns of the arithmetic sequence 2.5,4,5.5... must be taken for the sum to be greater than 200?

Jul 27, 2018

$16$ terms are required

#### Explanation:

${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$
where $a$ is the first term, $d$ is the difference between 2 adjacent terms and $n$ is the nth term

Looking at the sequence: $2.5 , 4 , 5.5 , \ldots$
$a = 2.5$
$d = 4 - 2.5 = 1.5$

Putting those two constants into the equation:
${S}_{n} = \frac{n}{2} \left[2 \left(2.5\right) + \left(n - 1\right) \left(1.5\right)\right]$
${S}_{n} = \frac{n}{2} \left[5 + 1.5 n - 1.5\right]$
${S}_{n} = \frac{n}{2} \left[1.5 n + 3.5\right]$

Now to find the number of terms needed for the sum to be greater than $200$,
ie ${S}_{n} > 200$

$\frac{n}{2} \left[1.5 n + 3.5\right] > 200$
($n$ is the unknown variable that we have to find)

$n \left[1.5 n + 3.5\right] > 200 \times 2$

$1.5 {n}^{2} + 3.5 n > 400$

$1.5 {n}^{2} + 3.5 n - 400 > 0$

$0.3 {n}^{2} + 0.7 n - 80 > 0$

Using the quadratic formula,

$n = \frac{- 0.7 \pm \sqrt{{0.7}^{2} - 4 \left(0.3\right) \left(- 80\right)}}{2 \times 0.3}$

$n = \frac{- 0.7 \pm \sqrt{0.49 + 96}}{0.6}$

$n = \frac{- 0.7 \pm \sqrt{96.49}}{0.6}$

$n = \frac{- 0.7 + \sqrt{96.49}}{0.6}$ only as $n > 0$ since $n$ is the number of terms

$n = 15.20488725$

That means that we must take a minimum of 16 terms in order for our sum to be greater than 200

To test, we sub $n = 15$ and $n = 16$ back into our equation ${S}_{n} = \frac{n}{2} \left[1.5 n + 3.5\right]$

If $n = 15$,
${S}_{15} = \frac{15}{2} \left[1.5 \times 15 + 3.5\right]$
${S}_{15} = 195$

If $n = 16$,
${S}_{16} = \frac{16}{2} \left[1.5 \times 16 + 3.5\right]$
${S}_{16} = 220$

Therefore, $16$ terms are required is correct