posted by Anonymous .
1) In the arithmetic progression, the 8th term is twice the 4th term and the 20th term is 40. Find the common difference and the sum of the terms from the 8th to the 20th term inclusive.
2) Find the least number of terms of the AP, 1+3+5...that are required to make a sum exceeding 4000.
Any help at all would be great
remember that t(n) = a+(n-1)d
t(8) = a+7d
t(4) = a+3d
it said: the 8th term is twice the 4th term ---> a+7d = 2(a+3d), simplify this
then it said: the 20th term is 40 ---> a+19d = 40
Solve these two equations.
2) YOu should know that
S(n) = n/2[2a + (n-1)d]
so you want n/2[2 + (n-1)(2)] > 4000
Th 4th term 0f an a p is 13 and the 10th term is 29 find the 16th term
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