Question

A2+a3+a4 =55 a3=? (Арифметическая прогрессия )

Answer 1

$a_{2}+a_{3} +a_{4} =55\\a_{1} +d +a_{1} +2d+a_{1} +3d = 55\\3a_{1} +6d= 55\: |:3\\a_{1} +2d = 18\dfrac{1}{3} \\a_{3}= 18\dfrac{1}{3}$

Answer 2

по определению члена арифметической прогрессии

a(n) = a(n-1) + d

a(n) = a(n+1) - d

a2= a3 - d

a4 = a3 + d

a2 + a3 + a4 = a3 - d + a3 + a3 + d = 3a3

3a3= 55

a3 = 55/3 =18 1/3

Answer 3

To find the value of a3 in an arithmetic progression, we need more information. In an arithmetic progression, each term (a) is obtained by adding a constant difference (d) to the previous term. The general formula for an arithmetic progression is:

a(n) = a1 + (n-1) * d

where: a(n) = the nth term of the arithmetic progression, a1 = the first term of the arithmetic progression, d = the common difference between consecutive terms, n = the position of the term in the arithmetic progression.

Since we have three consecutive terms A2, a3, and a4 with a sum of 55, we can use this information to find a3.

Let's assume a3 = x.

So, the sum of the three terms is:

A2 + a3 + a4 = 55

Substitute the formula for each term:

(a1 + d) + x + (a1 + 2d) = 55

Now, we need additional information to solve for x, a1, and d. For example, you could provide the value of a1 (the first term) or the common difference (d). Alternatively, if you have the values of any other terms or any other information related to the arithmetic progression, please include it in your question so that we can calculate the value of a3.