# First term is 5 and the common difference is 3. Starting with the first term, how do you find the number of terms that have a sum of 3925?

Apr 1, 2016

Use the method shown to find number of terms is $50$

#### Explanation:

The general term of an arithmetic sequence can be written:

${a}_{n} = a + d \left(n - 1\right)$

where $a$ is the initial term and $d$ the common difference.

Then:

${\sum}_{n = 1}^{N} {a}_{n} = N \cdot \frac{{a}_{1} + {a}_{N}}{2}$

$= N \frac{\left(a + a + d \left(N - 1\right)\right)}{2}$

$= a N + \frac{d}{2} \cdot N \left(N - 1\right)$

$= \frac{d}{2} {N}^{2} + \left(a - \frac{d}{2}\right) N$

In our example, $a = 5$, $d = 3$ and we want to solve

$3925 = {\sum}_{n = 1}^{N} {a}_{n}$

$= \frac{d}{2} {N}^{2} + \left(a - \frac{d}{2}\right) N$
$= \frac{3}{2} {N}^{2} + \left(5 - \frac{3}{2}\right) N$

$= \frac{3}{2} {N}^{2} + \frac{7}{2} N$

Multiply both ends by $2$ to get:

$3 {N}^{2} + 7 N = 7850$

Subtract $7850$ from both sides to get:

$3 {N}^{2} + 7 N - 7850 = 0$

Then use the quadratic formula to find:

$N = \frac{- 7 \pm \sqrt{{7}^{2} + 4 \cdot 3 \cdot 7850}}{6}$

$= \frac{- 7 \pm \sqrt{49 + 94200}}{6}$

$= \frac{- 7 \pm \sqrt{94249}}{6}$

$= \frac{- 7 \pm 307}{6}$

That is $N = - \frac{314}{6} = - \frac{157}{3}$ or $N = \frac{300}{6} = 50$

Discard the negative solution to find $N = 50$