Sum an arithmetic series with step 4

Question

Solve in your head. What is 3 + 7 + 11 + \dots + 99? (each term is 4 more than the last) Aim: under 15 seconds.

Accepted Answer

Count-times-average handles any constant step, you just need the number of terms. Terms go 3, 7, 11, \dots, 99 in steps of 4, so 	ext{count} = \frac{99 - 3}{4} + 1 = \frac{96}{4} + 1 = 24 + 1 = 25. The average of a arithmetic series is the average of its endpoints: \frac{3 + 99}{2} = 51. So 3 + 7 + \dots + 99 = 25 	imes 51 = 1275. Takeaway. For any evenly spaced series, count = \frac{	ext{last} - 	ext{first}}{	ext{step}} + 1, and sum = 	ext{count} 	imes \frac{	ext{first} + 	ext{last}}{2}. The step only affects the count; the endpoint-average rule never changes.

Sum an arithmetic series with step 4

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