Question

Сумма а7 и а4 равна 6 в арифметической прогрессии. а5 на 12 больше а2. найти а1 и а3

Answer 1

An=a1+d(n-1)
1)
a2+12=a5
a2=a1+d
a5=a1+4d
a1+d+12=a1+4d
a1-a1+d-4d=-12
-3d=-12
d=4

2)
a7+a4=6
a7=a1+6d
a4=a1+3d
a1+6d+a1+3d=6
2*a1+9d=6
2*a1+9*4=6
2*a1+36=6
2*a1=-30
a1=-15

3) a3=a1+2d
a3=-15+2*4=-15+8=-7

Ответ: а1=-15; а3=-7

Answer 2

Arithmetic Progression Problem

To solve the problem, we are given the following information: - The sum of a7 and a4 in an arithmetic progression is 6. - a5 is 12 more than a2.

Let's solve for a1 and a3.

Solution:

First, let's use the given information to form equations.

We know that the sum of a7 and a4 is 6: a7 + a4 = 6.

We also know that a5 is 12 more than a2: a5 = a2 + 12.

Now, we can use the formula for the nth term of an arithmetic progression: a(n) = a(1) + (n - 1)d, where a(n) is the nth term, a(1) is the first term, n is the term number, and d is the common difference.

We can also use the formula for the sum of the first n terms of an arithmetic progression: S(n) = (n/2)(a(1) + a(n)).

We can use these formulas to solve for a1 and a3.

Calculations:

Using the given information and the formulas for arithmetic progression, we can solve for a1 and a3.

From the given information, we have: 1. a5 = a2 + 12. 2. a7 + a4 = 6.

We can express a7 and a4 in terms of a1 and d: - a7 = a1 + 6d - a4 = a1 + 3d

Substituting these expressions into the equation a7 + a4 = 6, we get: a1 + 6d + a1 + 3d = 6 2a1 + 9d = 6 a1 + 3d = 3

Now, we can express a5 in terms of a1 and d: a5 = a1 + 4d

Substituting the given relationship a5 = a2 + 12, we get: a1 + 4d = a2 + 12

We can also express a3 in terms of a1 and d: a3 = a1 + 2d

Conclusion:

By solving the equations, we can find the values of a1 and a3.