# The first three terms of an arithmetic sequence are 2k+3, 5k-2 and 10k-15, how do you show that k =4?

May 17, 2016

Since the sequence is arithmetic, there is a number $d$ (the "common difference") with the property that $5 k - 2 = \left(2 k + 3\right) + d$ and $10 k - 15 = \left(5 k - 2\right) + d$. The first equation can be simplified to $3 k - d = 5$ and the second to $5 k - d = 13$. You can now subtract the first of these last two equations from the second to get $2 k = 8$, implying that $k = 4$.
Alternatively, you could have set $d = 3 k - 5 = 5 k - 13$ and solved for $k = 4$ that way instead of subtracting one equation from the other.
It's not necessary to find, but the common difference $d = 3 k - 5 = 3 \cdot 4 - 5 = 7$. The three terms in the sequence are 11, 18, 25.