Given a sequence with #a_1 = 2# and #a_3 = 6#, for what value of #n# is #a_n = 58# ?

Question

Lily Anderson · Accepted Answer

If this is an arithmetic sequence then #n=29#

We receive:

#a_1 = 2#

#a_3 = 6#

and want to find #n# such that #a_n = 58#

In the event that this is an arithmetic sequence, the following formula provides the general term:

#a_n = a+d(n-1)#

where #a# is the first term and #d# the common difference.

Next, we discover:

#4 = 6-2 = a_3-a_1 = (a+d(color(blue)(3)-1))-(a+d(color(blue)(1)-1)) = 2d#

So the common difference #d=2# and the first term #a=a_1 = 2#

Then:

#58 = a_n = a+d(n-1) = 2+2(n-1)#

Subtract #2# from both ends to get:

#56 = 2(n-1)#

Divide both sides by #2# to get:

#28 = n-1#

Add #1# to both sides to find:

#29 = n#

That is: #n = 29#

#color(white)()# Footnote

A sequence is any list of things. Without further specification, the answer could be any positive integral value of #n# apart from #1# or #3#.

Is it any other kind of sequence?

Given that it has terms #a_1 = 2#, #a_3 = 6#, #a_n = 58#, it cannot be a geometric sequence or a harmonic sequence.

It might be an odd variation of double Lucas sequence.

The Lucas sequence is like the Fibonacci sequence, but with seed values #L_0 = 2# and #L_1 = 1#. Each subsequent term is formed by adding the two previous ones. So, starting from #L_1#, it goes:

#1, 3, 4, 7, 11, 18, 29, 47,...#

Notice the terms #1#, #3# and #29#, i.e. half of #2#, #6# and #58#.

Thus, we could draft regulations:

#a_1 = 2#

#a_2 = 3#

#a_3 = 6#

#a_4 = 7#

#a_(n+4) = a_(n+2)+a_n" "# for #n >= 1#

This would provide the following order:

#2, 3, 6, 7, 8, 10, 14, 17, 22, 27, 36, 44, 58, 71, 94, 115, 152, 186,...#

Then #a_13 = 58#