• Nov 6th 2008, 07:03 AM
juliak
(Arithmetic Sequences)

How do I find K, given the consecutive arithmetic terms:
1. k, 8, k+11
2. k+2, 2k+3, 17

The only example we were given is:
-1 - (k+5) = (2k-1) - (-1)
-1 - k - 5 = 2k - 1 +1
-k -6 = 2k
-6 = 3k
k = -2

... which doesn't really apply, particularly through the first one.

I managed to figure out one:
31, k, 13
31-13=18
18/2=9
31-9=22
--> purely based on logic though, I don't know how to do the previous two.
• Nov 6th 2008, 07:26 AM
Soroban
Hello, juliak!

Quote:

How do I find $k$, given the consecutive arithmetic terms:

. . $1)\;\; k,\: 8,\: k+11\qquad\qquad 2)\;\;k+2,\:2k+3,\:17$

We are expected to know that with an arithmetic sequence,
. . consecutive terms have a common difference, $d.$
That is: . $d \:=\:a_2-a_1,\;\;d \:=\:a_3-a_2$

In the first problem, we have: . $a_1 \:=\:k,\;\;a_2\:=\:8,\;\;a_3 \:=\:k+11$

. . $d \:=\:8 - k$ .[1]

. . $d \:=\:(k+11) - 8 \quad\Rightarrow\quad d \:=\:k+3$ .[2]

Equate [1] and [2]: . $8 - k \:=\:k + 3 \quad\Rightarrow\quad -2k \:=\:-5$

Therefore: . $\boxed{k \:=\:\frac{5}{2}}$

Now try #2 yourself . . .