Question

Начиная с какого номера все члены последовательности (xn) будут не меньше заданного числа А: xn=4n-6, A=19?

Answer 1

Ответ:

n = 7

Пошаговое объяснение:

$a_{n} = a_{1} =+d(n-1) = -2 + 4(n-1)\\-2 + 4(n-1) > 19\\4(n-1) > 21\\n-1 > 5.25\\n > 6.25 ---- n = 7$

Answer 2

To find the starting term (number) in the sequence (xn) where all the members are not less than the given number A, which is A = 19, we need to set up the inequality:

xn >= A

Given that xn = 4n - 6, we can replace xn in the inequality:

4n - 6 >= 19

Now, let's solve for n:

4n >= 19 + 6 4n >= 25

Divide both sides by 4:

n >= 25 / 4 n >= 6.25

Since n must be a positive integer (we can't have fractions or decimal values for n in a sequence), the smallest integer value that satisfies the inequality is n = 7.

Therefore, starting from the 7th term (n = 7), all the members of the sequence (xn) will not be less than the given number A = 19.