Question

Найдите сумму двенадцати первых членов арифметической прогрессии, если а1= -10, d= 1,5 Срочно

Answer 1

$S_n=\frac{2a_1+d(n-1)}{2} *n$

$S_1_2=\frac{2*(-10)+1,5*(12-1)}{2} *12=\frac{-20+16,5}{1} *6=-21$

Answer 2

Calculating the Sum of the First Twelve Terms of an Arithmetic Progression

To find the sum of the first twelve terms of an arithmetic progression with \(a_1 = -10\) and \(d = 1.5\), we can use the formula:

Sum of first n terms of an arithmetic progression: \[S_n = \frac{n}{2} \times (2a_1 + (n-1)d)\]

Where: - \(S_n\) = sum of the first n terms - \(a_1\) = first term - \(d\) = common difference - \(n\) = number of terms

Calculation

Using the given values: - \(a_1 = -10\) - \(d = 1.5\) - \(n = 12\)

We can calculate the sum of the first twelve terms using the formula: \[S_{12} = \frac{12}{2} \times (2 \times (-10) + (12-1) \times 1.5)\]

\[S_{12} = 6 \times (-20 + 16.5)\]

\[S_{12} = 6 \times (-3.5)\]

\[S_{12} = -21\]

Therefore, the sum of the first twelve terms of the given arithmetic progression is -21.

[[SOURCE #1]]