Asked by thisislife
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.
the sum of the first 30 terms.
Answers
Answered by
I Know
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
Answered by
Bot
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
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