Math Problem Statement

rezolva 26,27,28,29,30,PE SCURT

rezolva 26,27,28,29,30,PE SCURT

Solution

Let's solve each of the problems (26 to 30) one by one briefly:

Problem 26:

Determine the sum of three terms of a geometric progression, knowing that the sum of the first two terms is 8, and the difference between the second and the third term is 4.

Let the terms of the geometric progression be aa, arar, and ar2ar^2. We are given:

  1. a+ar=8a + ar = 8a(1+r)=8a(1 + r) = 8
  2. ar2ar=4ar^2 - ar = 4ar(r1)=4ar(r - 1) = 4

From the first equation, solve for aa: a=81+ra = \frac{8}{1 + r}

Substitute this into the second equation: 8r(r1)1+r=4\frac{8r(r - 1)}{1 + r} = 4

Solve for rr, then substitute back to find aa. This will give the terms and their sum.


Problem 27:

Calculate the fifth term of an arithmetic progression, knowing that the first term is 7 and the second term is 9.

In an arithmetic progression, the nn-th term is given by: an=a1+(n1)da_n = a_1 + (n-1)d

Given:

  • a1=7a_1 = 7
  • a2=9a_2 = 9

We know: a2=a1+da_2 = a_1 + d Thus, 9=7+d9 = 7 + d, so d=2d = 2.

The fifth term: a5=7+(51)×2=7+8=15a_5 = 7 + (5-1) \times 2 = 7 + 8 = 15

So, the fifth term is 15.


Problem 28:

Determine the ratio of the geometric progression (bn)n1(b_n)_{n \geq 1}, knowing that b1=3b_1 = 3 and b2b1=3b_2 - b_1 = 3.

We are given:

  • b1=3b_1 = 3
  • b2b1=3b_2 - b_1 = 3

Since it's a geometric progression: b2=b1rb_2 = b_1 \cdot r

Thus: 3r3=33r - 3 = 3 3r=6r=23r = 6 \quad \Rightarrow \quad r = 2

So, the ratio of the geometric progression is r=2r = 2.


Problem 29:

Prove that for the sequence with the general term an=2n+3a_n = 2n + 3, the relation an+1an=2a_{n+1} - a_n = 2 holds for all nNn \in \mathbb{N}.

The general term is an=2n+3a_n = 2n + 3.

Now, calculate: an+1=2(n+1)+3=2n+2+3=2n+5a_{n+1} = 2(n+1) + 3 = 2n + 2 + 3 = 2n + 5

Now find the difference: an+1an=(2n+5)(2n+3)=2a_{n+1} - a_n = (2n + 5) - (2n + 3) = 2

Thus, the relation an+1an=2a_{n+1} - a_n = 2 is proven for all nNn \in \mathbb{N}.


Problem 30:

Determine the real number xx, knowing that the numbers x1x-1, x+1x+1, and 2x12x-1 are consecutive terms of an arithmetic progression.

In an arithmetic progression, the difference between consecutive terms is constant. So, we can write: (x+1)(x1)=(2x1)(x+1)(x+1) - (x-1) = (2x-1) - (x+1) Simplifying both sides: 2=2x22 = 2x - 2 Solving for xx: 2x=4x=22x = 4 \quad \Rightarrow \quad x = 2

Thus, the value of xx is 2.


Do you need more detailed steps for any of these problems?

Related Questions:

  1. How do you determine the common ratio in a geometric progression?
  2. What is the general term for an arithmetic progression?
  3. How can you prove relationships between consecutive terms in sequences?
  4. What is the formula for the sum of a finite geometric progression?
  5. How can you check if a sequence of numbers forms an arithmetic progression?

Tip:

Always double-check whether the progression is arithmetic or geometric before applying formulas!

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Math Problem Analysis

Mathematical Concepts

Geometric Progression
Arithmetic Progression
Sequences and Series
Algebra

Formulas

Sum of a geometric progression: S_n = a(1 - r^n) / (1 - r)
Arithmetic progression nth term: a_n = a_1 + (n-1) * d
Geometric progression nth term: b_n = b_1 * r^(n-1)
Difference between consecutive terms in a sequence

Theorems

Geometric progression common ratio
Arithmetic progression term difference

Suitable Grade Level

Grades 9-12