Question

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

Answer 1

Ответ:

aₙ - это последние число

a₁ - это первое число

d - это промежуток между чисел

n - это количество чисел

s - это общая сумма чисел

$a_{n} =a_{1} +d*(n-1)$

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

$91-1=3n-3$

$90+3=3n$

$93=3n$

$n=\frac{93}{3}$

$n=31$

$s=(a_{n} +a_1)*\frac{n}{2}$

$s=(91+1)*\frac{31}{2}$

$s=92*15.5$

$s=1426$

$1+4+7+10+...+91=1426$

Answer 2

It looks like you want to find the sum of an arithmetic series where the first term is 1, the common difference is 3, and the last term is 91. To find the sum of this series, you can use the formula for the sum of an arithmetic series:

$S = \frac{n}{2}[2a + (n-1)d]$

Where:

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

In this case:

• $a = 1$
• $d = 3$

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

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

Where:

• $T_n$ is the $n$th term.

We know that the last term is 91, so we can set $T_n = 91$ and solve for $n$:

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

Now, solve for $n$:

$91 = 1 + 3n - 3$

Combine like terms:

$91 = 3n - 2$

Add 2 to both sides:

$93 = 3n$

Now, divide by 3:

$n = \frac{93}{3} = 31$

So, there are 31 terms in the series.

Now that we know $n$, we can plug the values into the sum formula:

$S = \frac{31}{2}[2(1) + (31-1)(3)]$

Simplify further:

$S = \frac{31}{2}[2 + 30(3)]$

$S = \frac{31}{2}[2 + 90]$

$S = \frac{31}{2}[92]$

Now, calculate the final sum:

$S = \frac{31 \cdot 92}{2}$

$S = 31 \cdot 46$

$S = 1426$

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