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Math is fun arithmetic sequence
a = 31
d = 1
Tn = a + (n-1) d
here (n-1) d = 230 - 31 = n-1
so n = 200
n =
Find the sum of 31 + 32 + 33 + ... + 230. Show your work.
3 answers
230-31 = 199
so there are 200 values, forming 100 pairs of numbers which add up to 230+31 = 261
the sum is 100*261 = 26,100
so there are 200 values, forming 100 pairs of numbers which add up to 230+31 = 261
the sum is 100*261 = 26,100
sum from a to a + (n-1)d = (n/2)[ 2 a + (n-1)d ]
= (200/2)(2*31 + 199) = 100 *261 = 26100
= (200/2)(2*31 + 199) = 100 *261 = 26100
1 answer