Question

Sn=-441 for the series 19 + 15 + 11 ... tn. Determine the value of n.

Answer 1

Answer:

sn = d/2·n^2 + (a - d/2)·n

sn = (-4)/2·n^2 + (19 - (-4)/2)·n = -441

- 2·n^2 + 21·n = -441

- 2·n^2 + 21·n + 441 = 0

n^2 - 21/2·n - 441/2 = 0

use pq-Formula

n = -10.5 ∨ n = 21

only n = 21 makes sense. So n = 21 is the correct answer.

Answer 2

Answer:

The value of n is 21

Step-by-step explanation:

The series is 19 , 15 , 11 , ............

It has a constant difference between each two consecutive terms, so it is an arithmetic series.

The rule of the sum of n terms in the arithmetic series is:

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

Where a is the first term of the series , d is the difference between each two consecutive terms and n is the number of the terms in this series

So from the series above:

a = 19 , d = 15 - 19 = -4 ( it must be second term - first term ) , $S_{n}=-441$

$-441=\frac{n}{2}[2(19)+(n-1)(-4)]$

$-441=\frac{n}{2}[38+(-4n+4)]$

$-441=\frac{n}{2}[38-4n+4]$

$-441=\frac{n}{2}[42-4n]$

$-441=21n-2n^{2}$

$2n^{2}-21n-441=0$ ⇒Use factorization

(2n + 21) (n - 21) = 0

2n + 21 = 0 ⇒ 2n = -21 ⇒ n = -21/2⇒Not accepted n must be positive integer

n - 21 = 0⇒ n = 21

∴-441 is the sum of 21 terms in this series