Mark M. answered • 09/27/18
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Mathematics Teacher - NCLB Highly Qualified
a5 + a8 + a11 = 10
a5 + a5 + 3d + a5 + 6d = 10
3a5 + 9d = 10
a7 + a10 + a13 = 12
a5 + 3d + a5 + 6d + a5 + 9d = 12
3a5 + 18d = d
Now solve for d and determine k such that ak = 11
Al P.
Yes, k = 31.
For completeness:
The two equations are 3a5+9d = 10 and 3a5+15d = 12 ==> d=1/3 and a5=7/3
ak = 11 = a5 + (k-5)d
11 = 7/3 + (k-5)(1/3)
33 = 7 + k - 5 —> k = 31
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09/28/18
Bader A.
09/27/18