Sum of first 30 terms = (30/2)[2a + (30 - 1)d]
a = 6th term = 13
d = common difference = (3420 - 13)/(40 - 1) = 83.5
Sum of first 30 terms = (30/2)[2(13) + (30 - 1)(83.5)]
Sum of first 30 terms = (30/2)[26 + 2452.5]
Sum of first 30 terms = (30/2)(2478.5)
Sum of first 30 terms = 14673.5
show your work The 6th term of an arithmetic series is 13, the sum of the first 40 terms is 3420. Find
the sum of the first 30 terms.
2 answers
Totally botched by the bot !!!
correct solution:
a + 5d = 13, or a = 13-5d
(40/2)(2a + 39d) = 3420
20(2a + 39d) = 3420
2a + 39d = 171
2(13-5d) + 39d = 171
solving this for d ...
d = 5
then a = 13-5(5) = -12
sum(30) = 15(-24 + 29(5)) = 1815
correct solution:
a + 5d = 13, or a = 13-5d
(40/2)(2a + 39d) = 3420
20(2a + 39d) = 3420
2a + 39d = 171
2(13-5d) + 39d = 171
solving this for d ...
d = 5
then a = 13-5(5) = -12
sum(30) = 15(-24 + 29(5)) = 1815