Question

A1=7 d=2 a12-? S15-? Арифметическая прогрессия.

Answer 1

а12=7+2*11=15

s15=(2*7+(15-1))*15/2=(2*7+2*14)*15/2=42*7,5=315

ну как-то так))

Answer 2

To find the values of a12 and S15 in the arithmetic progression, we need to use the formulas for the nth term (a_n) and the sum of the first n terms (S_n) of an arithmetic progression.

In an arithmetic progression, the nth term (a_n) is given by:

a_n = a1 + (n - 1) * d

where: a_n is the nth term, a1 is the first term of the progression (A1 in this case), n is the term number whose value we want to find, d is the common difference between consecutive terms (given as 2 in this case).

The sum of the first n terms (S_n) in an arithmetic progression is given by the formula:

S_n = (n / 2) * (a1 + a_n)

where: S_n is the sum of the first n terms, n is the number of terms, a1 is the first term, a_n is the nth term.

Now, let's find a12 and S15:

Given: A1 = 7 (the first term) d = 2 (common difference)

1. Finding a12: a12 = a1 + (12 - 1) * d a12 = 7 + 11 * 2 a12 = 7 + 22 a12 = 29

So, the 12th term (a12) in the arithmetic progression is 29.

2. Finding S15: S15 = (15 / 2) * (a1 + a15) S15 = (15 / 2) * (7 + a15)

Now, we need to find a15. Using the same formula as before: a15 = a1 + (15 - 1) * d a15 = 7 + 14 * 2 a15 = 7 + 28 a15 = 35

Now, we can substitute a15 back into the equation for S15: S15 = (15 / 2) * (7 + 35) S15 = (15 / 2) * 42 S15 = 15 * 21 S15 = 315

So, the sum of the first 15 terms (S15) in the arithmetic progression is 315.