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

Question

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

Answer 1

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!!!

Answer 2

a1+a2+a3+a4+a5+a6

Answer 3

a1+a2+a3+a4+a5+a6=
9, 6, 3, 0, -3, -6, -9