Find the sum. sum_(k=1)^(40)(25-2k)=

Find Arithmetic Series Sum: We need to find the sum of the arithmetic series where the first term a_1 is 25 - 2(1) = 23, the common difference d is -2, and the number of terms n is 40.Use Sum Formula: To find the sum of an arithmetic series, we use the formula S_n = 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.Calculate 40th Term: First, we need to find the 40th term, a_40. Since the common difference d is -2, we can calculate a_40 as a_1 + (n-1)d = 23 + (40-1)(-2).Find a_40: Calculating a_40 gives us 23 + 39(-2) = 23 - 78 = -55.Plug Values into Formula: Now we have a_1 = 23 and a_40 = -55. We can plug these values into the sum formula: S_40 = 40/2 * (23 - 55).Simplify Sum Formula: Simplifying the sum formula gives us S_40 = 20 * (-32).Calculate Final Sum: Calculating S_40 gives us 20 * (-32) = -640.

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Grade 8

Negative Exponents

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Q. Find the sum.

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\sum_{k=1}^{40}(25-2 k)=

Find Arithmetic Series Sum: We need to find the sum of the arithmetic series where the first term $a_1$ is $25 - 2(1) = 23$ , the common difference $d$ is $-2$ , and the number of terms $n$ is $40$ .
Use Sum Formula: To find the sum of an arithmetic series, we use the formula $S_n = \frac{n}{2} \times (a_1 + a_n)$ , where $S_n$ is the sum of the first $n$ terms, $a_1$ is the first term, and $a_n$ is the $n$ th term.
Calculate $40$ th Term: First, we need to find the $40$ th term, $a_{40}$ . Since the common difference $d$ is $-2$ , we can calculate $a_{40}$ as $a_1 + (n-1)d = 23 + (40-1)(-2)$ .
Find $a_{40}$ : Calculating $a_{40}$ gives us $23 + 39(-2) = 23 - 78 = -55$ .
Plug Values into Formula: Now we have $a_1 = 23$ and $a_{40} = -55$ . We can plug these values into the sum formula: $S_{40} = \frac{40}{2} \times (23 - 55)$ .
Simplify Sum Formula: Simplifying the sum formula gives us $S_{40} = 20 \times (-32)$ .
Calculate Final Sum: Calculating $S_{40}$ gives us $20 \times (-32) = -640$ .