Posted by **Jared** on Wednesday, September 30, 2009 at 8:56pm.

suppose that a1, a2, a3, ..., ak forms an arithmetic sequence. If a5+a8+a11=10, a7+a10+a13=12, and ak=11, find k

note: the numbers and the k after the a's are subscript but i don't know how to type that

- advanced math -
**Reiny**, Wednesday, September 30, 2009 at 9:08pm
from your first equation :

a+4d + a+7d + a+10d = 10

3a + 21d = 10 (#1)

from your second equation :

a+6d + a+9d + a+12d = 12

3a 27d = 12 (#2)

subtract #2 - #1

6d = 2

d = 1/3

back in #1, 3a + 21(1/3) = 10

a = 1

then a+(kn-1)d = 11

1 + (k-1)(1/3) = 11

3 + k-1 = 33

**k** = 31

check : term(31) = a+30d

= 1 + 30(1/3)

= 1 + 10 = 11 Yea!!!

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