Question

Compute the sums below. (Assume that the terms in the first sum are consecutive terms of an arithmetic sequence.) $3 + 1 + (-1) + \dots + (-249) =$ $\sum_{j=1}^{100} (-5j + 9) = $

          Compute the sums below. (Assume that the terms in the first sum are consecutive terms of an arithmetic sequence.)
$3 + 1 + (-1) + \dots + (-249) =$
$\sum_{j=1}^{100} (-5j + 9) = $

Compute the sums below. (Assume that the terms in the first sum are consecutive terms of an arithmetic sequence.)
3 + 1 + (-1) + … + (-249) =
∑j=1^100 (-5j + 9) =

Added by Erika A.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Zack A

11/23/2023

Step 1

The first sum starts at 3 and ends at -249, with a common difference of -2. To find the number of terms, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n-1)d, where an is the nth term, a1 is the first term, n is the number of terms, Show more…

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Compute the sums below. (Assume that the terms in the first sum are consecutive terms of an arithmetic sequence.) 3+1+(-1)+dots+(-249)= sum_(j=1)^(100) (-5j+9)= Compute the sums below. (Assume that the terms in the first sum are consecutive terms of an arithmetic sequence.) 3+1+-1+.+-249= X 100 (-5+- j=1