SOLUTION: the 4th term of an A.P is 15 and the 9th term is 35. Find the fifteenth term?

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Question 1016403: the 4th term of an A.P is 15 and the 9th term is 35. Find the fifteenth term?
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
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the 4th term of an A.P is 15 and the 9th term is 35. Find the fifteenth term?
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Apply the formula a%5Bn%5D = a%5B1%5D+%2B+%28n-1%29%2Ad for the n-th term of an AP. You will have

a%5B4%5D = a%5B1%5D + 3%2Ad = 15.   (1)

a%5B9%5D = a%5B1%5D + 8%2Ad = 35.   (2)

Take the difference (2) - (1). You will have

8d - 3d = 35 - 15  --->  5d = 20  --->  d = 20%2F5 = 4.

Thus we just found the common difference.

Now, a%5B15%5D = a%5B1%5D+%2B+14%2Ad = %28a%5B1%5D+%2B+8%2Ad%29 + 6d = a%5B9%5D + 6d = 35 + 6*4 = 35 + 24 = 59.

Answer. a%5B15%5D = 59.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

the 4th term of an A.P is 15 and the 9th term is 35. Find the fifteenth term?
4th term
a%5Bn%5D+=+a%5B1%5D+%2B+%28n+-+1%29d
a%5B4%5D+=+a%5B1%5D+%2B+%284+-+1%29d
15+=+a%5B1%5D+%2B+3d ------- eq (i)
9th term
a%5Bn%5D+=+a%5B1%5D+%2B+%28n+-+1%29d
a%5B9%5D+=+a%5B1%5D+%2B+%289+-+1%29d
35+=+a%5B1%5D+%2B+8d ------- eq (ii)
- 20 = - 5d ------ Subtracting eq (ii) from eq (i)
d, or common difference = %28-+20%29%2F%28-+5%29, or 4
35+=+a%5B1%5D+%2B+8%284%29 ------- Substituting 4 for d in eq (i)
35+=+a%5B1%5D+%2B+32
matrix%281%2C8%2Ca%5B1%5D%2C+or%2C+1%5Est%2C+term%2C+%22=%22%2C+35+-+32%2C+or%2C+3%29
a%5Bn%5D+=+a%5B1%5D+%2B+%28n+-+1%29d
a%5B15%5D+=+3+%2B+%2815+-+1%294 ------- Substituting 15 for n, 3 for a%5B1%5D, and 4 for d
a%5B15%5D+=+3+%2B+14%284%29