Mathematics: Sn=-441 for the series 19 + 15 + 11 tn. determine the value of n. (i’ve gotten as for as -441 = 21n - 2n^2, but i can’t seem to understand how to get to the next

Log3(7) + log3(x)Terms that are multiplied inside one log become added when you separate them where each term gets its own log....

Sn=-441 for the series 19 + 15 + 11 tn. determine - id4411915115025621, Serenitybella

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18.01.2022

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

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