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Pre - calculus help please!!?

0 votes

The fifth and 50th terms of an arithmetic sequence are 6 and 33, respectively. Find the sum of the first 10 terms. 
please show any work.

asked Dec 8, 2014 in PRECALCULUS by anonymous
reshown Dec 8, 2014 by yamin_math

1 Answer

0 votes

Formula for Sum of n terms Sn = n/2[2a + (n - 1)d]

where a is first term and d is common difference.

Now calculate the a, d values in the sequence.

nth term in a arithmetic sequence = a + (n - 1)d

5th term in a arithmetic sequence = a + 4d

50th term in a arithmetic sequence = a + 49d

In this case a + 4d = 6 ---> (1)

a + 49d = 33 --> (2)

 

Now solve the equations (1) and (2).

To eliminate the a variable, subtract equation(1) from equation (2).

(a + 49d) - (a + 4d) = 33 - 6

45d = 27

d = 27/45

d = 3/5

Substitute d = 3/5 in equation(1).

a + 4(3/5) = 6

a = 6 - (12/5)

a = (30 - 12)/5

a = 18/5

Substitute n = 10, a = 18/5 and d = 3/5 in Sn.

S10 = (10/2)[2(18/5) + (10 - 1)(3/5)]

= 5[ (36/5) + (27/5)]

= 63

Sum of the first ten terms in the sequence = 63.

answered Dec 8, 2014 by david Expert

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