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Arithmetic Sequence Help?

0 votes
Can some one show me the steps to solve this arithmetic sequence problem?

Given:
6th term + 11th term = 51
2nd term + 8th term = 32

Required:
First term and common difference

Thanks
asked Feb 21, 2013 in ALGEBRA 2 by andrew Scholar

2 Answers

0 votes

Arithmetic Sequence in nth term is [a + (n-1)d]

Given: 6th term + 11th term = 51 and 2nd term + 8th term = 32

So, 17th term = 51 and 10th term = 32

17th term = a + 16d = 51-------->(1)

10th term = a + 9d = 32---------->(2)

Solve equation (1) and euqation (2)

a + 16d = 51

a + 9d = 32

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

7d = 19

Divide each side by 7.

d = 19/7.

Substituting d = 19/7 in the equation (2)

a + 9(19/7) = 32.

a + 171/7 = 32.

Subtract 171/7 from each side.

a = 32 - 171/7

Rewrite the expression with common denominator.

a = [224 - 171]/7

a = 53/7

Therefore first term is a = 53/7 and common difference is d = 19/7.

answered Feb 21, 2013 by britally Apprentice

First term = 36/7 and common difference is 19/7.

0 votes

Given: 6th term + 11th term = 51 and 2nd term + 8th term = 32

  • In Arithmetic Sequence n th term is [a + (n - 1)d ]

Where a = first term and d is common difference.

6 th term = a + 5d

11 th term = a + 10d

2 nd term = a + d

8 th term  = a + 7d

  • 6th term + 11th term = 51

a + 5d + a + 10d = 51

2a + 15d = 51 ------> (1)

2nd term + 8th term = 32

a + d + a + 7d = 32

2a + 8d = 32 ------> (2)

  • Solve equations (1) and (2).

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

(2a + 15d ) - (2a + 8d ) = 51 - 32

2a + 15d - 2a - 8d = 19

7d = 19

d = 19/7

Substitute d value in equation (2).

2a + 8(19/7) = 32

2a + 152/7 = 32

2a = 32 - 152/7

2a = (224 - 152)/7

2a = 72/7

a = 72/2*7

a = 36/7

First term 36/7 and common difference 19/7.

answered Jun 21, 2014 by david Expert

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