Asked by Fasi
In an arithmetic series, the sum of the 5th to the 8th terms inclusive is 114, whilst the sum of the 12th to 15th terms inclusive is 198. What is the 21st term?
I have got 99=2a +3d and 57= 2a+3d Im not sure what to do now
I have got 99=2a +3d and 57= 2a+3d Im not sure what to do now
Answers
Answered by
Steve
clearly both those equations cannot be true, since 99≠57.
The sum of T5 to T8 is
S8 - S4
= 8/2(2a+7d) - 4/2(2a+3d)
= 4a+22d = 114
Similarly,
S15-S11 = 15/2(2a+14d) - 11/2(2a+10d)
= 4a+50d = 198
Subtract those two to get 28d=84, or d=3
So, a=12
T21 = 12 + 20*3 = 72
The sum of T5 to T8 is
S8 - S4
= 8/2(2a+7d) - 4/2(2a+3d)
= 4a+22d = 114
Similarly,
S15-S11 = 15/2(2a+14d) - 11/2(2a+10d)
= 4a+50d = 198
Subtract those two to get 28d=84, or d=3
So, a=12
T21 = 12 + 20*3 = 72
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