Question

Find how many terms of the following series are needed to make the given sum: 5+8+11+14+...=670?

Answer 1

$670 = n/2[2*5+(n-1)*3]$
$670 = n/2[10+(n-1)*3]$
$1340 = n[10+3n-3]$
$1340 = n[7+3n]$
$1340 = 7n+3n^{2}$
$3n^{2}+7n-1340 = 0$
$n^{2}+\frac{7}{3}n-\frac{1340}{3} = 0$
$n^{2}+\frac{7}{3}n = \frac{1340}{3}$
$n^{2}+\frac{7}{3}n+\frac{49}{36}$ $=$ $\frac{1340}{3}+\frac{49}{36}$
$n^{2}+\frac{7}{3}n+\frac{49}{36} = \frac{16129}{36}$
$(n+\frac{7}{6})^2 = \frac{16129}{36}$
$n+\frac{7}{6} = \sqrt{\frac{16129}{36}}$
$n = -\frac{7}{6}+\sqrt{\frac{16129}{36}}$
$n = -1.167 + 21.167$
$n = 20$

Answer 2

You need 16 more terms not 17 because you already have 4 terms in the series so if you add those 4 terms to the other 16 terms to the series you get a total of 20 terms in total. I hope this helped.