Probably the simplest approach is to look for a pattern. If you let #a_{1}=10# be the first term, #a_{2}=14# be the second term, #a_{3}=18# be the third term, #a_{4}=22# be the fourth term, etc..., you should note that #a_{2}=a_{1}+4=10+4#, #a_{3}=a_{2}+4=a_{1}+8=a_{1}+2*4=10+2*4#, #a_{4}=a_{3}+4=a_{1}+2*4+4=a_{1}+3*4=10+3*4#, etc...

From this, I hope you see that the pattern is #a_{n}=10+(n-1)*4#. This is the formula for the #n^{th}# term. It can be simplified to #a_{n}=6+4n# (for #n=1,2,3,4,...#)

You can check that this works. For instance, #a_{5}=6+4*5=26# (you made a typo when you wrote 24) and #a_{6}=6+4*6=30#. You can even find the millionth term, or any other, this way: #a_{1000000}=4000006#.

This is an example of an *arithmetic sequence* . In general, an arithmetic sequence takes the form #a_{1}, a_{2}=a_{1}+d, a_{3}=a_{2}+d=a_{1}+2d,a_{4}=a_{3}+d=a_{1}+3d,...# for some numbers #a_{1}# (the first term) and #d# (the common difference). The pattern shows that #a_{n}=a_{1}+(n-1)d=(a_{1}-d)+dn# for #n=1,2,3,...# This is a (discrete) linear function of #n#.

For extra fun, we can find the sum of the first #N# terms of such an arithmetic sequence can be found by using the fact that #1+2+3+\cdots+(N-1)=((N-1)N)/2# (check this!):

#a_{1}+a_{2}+a_{3}+\cdots+a_{N}#

#=a_{1}+(a_{1}+d)+(a_{1}+2d)+\cdots+(a_{1}+(N-1)d)#

#=(a_{1}+a_{1}+a_{1}+\cdots+a_{1})+(d+2d+3d+\cdots+(N-1)d)#

#=a_{1}N+d*((N-1)N)/2=d/2 N^2+(a_{1}-d/2)N#

#=N/2(2a_{1}+d(N-1))#

Since #a_{N}=a_{1}+d(N-1)#, this can also be written as #N/2(a_{1}+a_{N})=N*(a_{1}+a_{N})/2# or #N*(\mbox{average of the first and last terms)#