How do you find the number of terms n given the sum $s_n=1661$ of the series $2+9+16+23+30+...$ ?

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Answer 1 · 2016-11-05T22:12:16

The series has $22$ terms.

We know that the common difference is $7$ , that the sum of n terms is 1661 and that the first term is $2$ .

So, using the formula $s_n = n/2(2a + (n - 1)d)$ , we can solve for $n$ .

$1661 = n/2(2(2) + (n - 1)7)$

$1661 = n/2(4 + 7n - 7)$

$1661 = n/2(7n - 3)$

$1661 = (7n^2)/2 - (3n)/2$

$3322 = 7n^2 - 3n$

$0 = 7n^2 - 3n - 3322$

$n = (-(-3) +- sqrt(-3^2 - 4 xx 7 xx -3322))/(2 xx 7)$

$n = (3 +- 305)/14$

$n = -302/14 and 308/14->"a negative answer is impossible"$

$n = 22$

How do you find the number of terms n given the sum s_n=1661s_n=1661 of the series 2+9+16+23+30+...2+9+16+23+30+...?