Question

1. Найдите сумму первых семи членов прогрессии 1; 1,8;2,6... 2. В арифметической прогрессии \ a n \ Найдите S_{6} , 1H * a_{1} = 1, 2 , a_{4} = 1, 8 S n - сумма первых л её членов. 3. В арифметической прогрессии \ a n \ S n - сумма первых п её членов. Найдите разность прогрессии, если a_{5} = 21 , S_{10} = 235 .Всё с решениямиПрошу срочно​

Answer 1

Ответ:

Объяснение:

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Answer 2

Arithmetic Progression Problems

1. Find the sum of the first seven terms of the progression 1; 1.8; 2.6...

To find the sum of the first seven terms of the given arithmetic progression, we can use the formula for the sum of the first n terms of an arithmetic progression: \( S_n = \frac{n}{2}(a_1 + a_n) \), where \( S_n \) is the sum of the first n terms, \( a_1 \) is the first term, and \( a_n \) is the nth term.

Using the given progression, we can calculate the sum as follows: \[ S_7 = \frac{7}{2}(1 + 2.6) = \frac{7}{2}(3.6) = 12.6 \]

Therefore, the sum of the first seven terms of the given progression is 12.6.

2. In an arithmetic progression with \( a_1 = 1 \), \( a_4 = 1.8 \), find \( S_6 \).

Given \( a_1 = 1 \) and \( a_4 = 1.8 \), we can find the common difference (\( d \)) using the formula \( a_n = a_1 + (n-1)d \). Substituting the given values, we get: \[ 1.8 = 1 + 3d \] \[ 3d = 0.8 \] \[ d = \frac{0.8}{3} = 0.2667 \]

Now, we can find \( S_6 \) using the formula \( S_n = \frac{n}{2}(a_1 + a_n) \): \[ S_6 = \frac{6}{2}(2 \times 1 + 5 \times 0.2667) = 3(2 + 1.3335) = 9.0005 \]

Therefore, the sum of the first six terms of the given progression is approximately 9.0005.

3. In an arithmetic progression with \( a_5 = 21 \) and \( S_{10} = 235 \), find the difference of the progression.

Given \( a_5 = 21 \) and \( S_{10} = 235 \), we can find the common difference (\( d \)) using the formula \( S_n = \frac{n}{2}(2a_1 + (n-1)d) \). Substituting the given values, we get: \[ 235 = \frac{10}{2}(2 \times a_1 + 9d) \] \[ 235 = 5(2a_1 + 9d) \]

We also know that \( a_5 = a_1 + 4d = 21 \). Using this, we can find \( a_1 \) in terms of \( d \): \[ a_1 = 21 - 4d \]

Substituting this into the equation for \( S_{10} \), we get: \[ 235 = 5(2(21 - 4d) + 9d) \] \[ 235 = 5(42 - 8d + 9d) \] \[ 235 = 5(42 + d) \] \[ 235 = 210 + 5d \] \[ 5d = 25 \] \[ d = 5 \]

Therefore, the difference of the given progression is 5.