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# What is the fifth term of an arithmetic sequence whose first term is 40 and whose seventh term is 121?

A.80

B.64

C.94

D.85

ALgebra problem

### 2 Answers

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- PuzzlingLv 74 weeks ago
Going from the 1st term to the 7th term is *6* steps. And you go up by 81.

81/6

= 27/2

= 13.5

So the step size (common difference) is 13.5

Going backwards 2 steps from 121, we would go back 2*13.5 = 27

121 - 27

= 94

Answer:

C. 94

- Wayne DeguManLv 74 weeks ago
nth term is a + (n - 1)d...where a is the 1st term and d is the common difference.

so, 7th term is 40 + 6d = 121

Hence, 6d = 81

=> d = 27/2

Then, nth term is 40 + (27/2)(n - 1)

i.e. 53/2 + 27n/2 => (53 + 27n)/2

so, 5th term is (53 + 135)/2 = 94

:)>

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