How to find the fifteenth term of the indicated series?

a5 + a10

= a + 4d + a + 9d

= 2a + 13d

= 100

9a + 36d = 270

18a + 117d = 900

18a + 72d = 540

45d = 360

d = 8

2a + 13*8 = 100

2a = 100 -104

a = -2

-2, 6, 14, 22, ...

an = 2(4n - 5)

a15 = 2(4*15 - 5)

The 15th term of the indicated

sequence is 110.

Let the nth term of the sequence be a(n) = a(1) + (n - 1)*d, where d is the common

difference. We are given that

a(5) + a(10) = [a(1) + (5 - 1)*d] + [a(1) + (10 - 1)*d] = 100 ----> 2*a(1) + 13*d = 100.

Now the sum of the first n terms of an arithmetic sequence is

S(n) = (n/2)*(2*a(1) + (n - 1)*d). We are given that

S(9) = (9/2)*(2*a(1) + (9 - 1)*d) = 9*a(1) + 36*d = 270.

So we have two equations and two unknowns. So multiply 2*a(1) + 13*d = 100

by 9 and multiply 9*a(1) + 36*d = 270 by 2 and then subtract:

18*a(1) + 117*d = 900

18*a(1) + 72*d = 540

------------------------------

45*d = 360 ----> d = 8, and so a(1) = (1/2)*(-13*8 + 100) = -2.

Thus a(15) = a(1) + (n - 1)*d = -2 + (15 - 1)*8 = 110.

For an arithmetic series whose first term is a and common difference is d,

the nth term = a + (n - 1)d

the sum of the first n terms is (n/2)(2a + [n - 1]d)

So, for this series, the 5th term = a + 4d; the 10th term = a + 9d

That gives a + 4d + a + 9d = 2a + 13d = 100

We also know that (9/2)(2a + 8d) = 9a + 36d = 270

So now we have a system of equations:

9a + 36d = 270

2a + 13d = 100 Solve by elimination: multiply top by 2, bottom by - 9:

18a + 72d = 540

- 18a - 117d = - 900

So, - 45d = - 360

d = 8. Now we need the first term:

2a + 13d = 100 becomes 2a + 104 = 100, so a = - 2

Now we can compute the 15th term:

- 2 + 14*8 = 110

n/2 (2a + (n-1)d) = 270

9/2 (2a + 8d) = 270

2a + 8d = 60

a + 4d + a + 9d = 100

2a + 13d = 100

2a + 8d = 60 (solving both the equations simutanously)

5 d = 40

d = 8

2a + 64 = 60

a = -2

15th term = a + 14d

-2 + 14*8 = 110

Tn=a+(n-1)d......................use this and your mind