Question

1+4+7+10+...+91 pls help

Answer 1

Ответ:

13+16+19+22+25+28+31+34+37+40+43+46+49+52+55+58+61+64+67+70+73+76+79+82+85+88+91

Answer 2

Ответ:

10-13-16-19-22-25-28-31-34-37-40-43-46-49-52-55-58-61-64-67-70-73-76-79-82-85-88-91

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Answer 3

To find the sum of the arithmetic sequence 1, 4, 7, 10, ..., 91, you can use the formula for the sum of an arithmetic series. The formula is:

$S_n = \frac{n}{2}(2a + (n-1)d)$

Where:

• $S_n$ is the sum of the series.
• $n$ is the number of terms in the series.
• $a$ is the first term in the series.
• $d$ is the common difference between terms.

In this case, the first term ($a$) is 1, the common difference ($d$) is 4 - 1 = 3, and we need to find the number of terms ($n$).

To find $n$, we can use the formula for the $n$th term of an arithmetic sequence:

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

Since the last term of the series is 91, we can set $a_n$ to 91 and solve for $n$:

$91 = 1 + (n-1)3$

Now, solve for $n$:

$91 = 1 + 3n - 3$

Add 3 to both sides:

$91 + 3 = 3n$

$94 = 3n$

Now, divide by 3:

$n = \frac{94}{3}$

So, there are approximately $n \approx 31.33$ terms in the series. Since you can't have a fraction of a term, you should use 31 terms.

Now, you can plug these values into the sum formula:

$S_{31} = \frac{31}{2}(2 \cdot 1 + (31-1) \cdot 3)$

Calculate the sum:

$S_{31} = \frac{31}{2}(2 + 90)$

$S_{31} = \frac{31}{2}(92)$

Now, multiply:

$S_{31} = 31 \cdot 46$

$S_{31} = 1426$

So, the sum of the series 1 + 4 + 7 + 10 + ... + 91 is 1426.