ASSIGNMENTFind the sum of each arithmetic sequence.1. 2, 5, 8, to 8 terms2. 3,5,7. . to 31 terms3. Find the sum of the first ten terms of thearithmetic series 4+10+16+.4. Find the sum of all multiples of 3 between 1and 100...​

Step-by-step explanation:Here are the solutions to the given arithmetic sequence problems:1. 2, 5, 8, ... to 8 terms * First term (a₁) = 2 * Common difference (d) = 5 - 2 = 3 * Number of terms (n) = 8Using the formula for the sum of an arithmetic sequence:Sₙ = n/2 [2a₁ + (n-1)d]S₈ = 8/2 [2(2) + (8-1)3]S₈ = 4 [4 + 21]S₈ = 4 * 25S₈ = 100Therefore, the sum of the first 8 terms of the sequence 2, 5, 8, ... is 100.2. 3, 5, 7, ... to 31 terms * First term (a₁) = 3 * Common difference (d) = 5 - 3 = 2 * Number of terms (n) = 31Using the formula for the sum of an arithmetic sequence:Sₙ = n/2 [2a₁ + (n-1)d]S₃₁ = 31/2 [2(3) + (31-1)2]S₃₁ = 31/2 [6 + 60]S₃₁ = 31/2 * 66S₃₁ = 1023Therefore, the sum of the first 31 terms of the sequence 3, 5, 7, ... is 1023.3. Find the sum of the first ten terms of the arithmetic series 4+10+16+... * First term (a₁) = 4 * Common difference (d) = 10 - 4 = 6 * Number of terms (n) = 10Using the formula for the sum of an arithmetic sequence:Sₙ = n/2 [2a₁ + (n-1)d]S₁₀ = 10/2 [2(4) + (10-1)6]S₁₀ = 5 [8 + 54]S₁₀ = 5 * 62S₁₀ = 310Therefore, the sum of the first 10 terms of the sequence 4, 10, 16, ... is 310.4. Find the sum of all multiples of 3 between 1 and 100 * First multiple of 3 (a₁) = 3 * Last multiple of 3 less than 100 is 99 * Common difference (d) = 3 * To find the number of terms (n), we can use the formula:aₙ = a₁ + (n-1)d99 = 3 + (n-1)396 = 3(n-1)32 = n-1n = 33Now, using the formula for the sum of an arithmetic sequence:Sₙ = n/2 [a₁ + aₙ]S₃₃ = 33/2 [3 + 99]S₃₃ = 33/2 * 102S₃₃ = 1683Therefore, the sum of all multiples of 3 between 1 and 100 is 1683.

Answered by xiancarlo20111120 | 2024-09-05