Arithmetic sequence

The terms a_1,a_2,a_3 form an arithmetic sequence whose sum is 18. The terms a_1+1,a_2,a_3+2 in that order, form a geometric sequence. Find the sum of all possible values for a_1.
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
Let d represent the common difference of the arithmetic sequence
a_1=a_1
a_2=a_1+d
a_3=a_1+2d
a_1+a_2+a_3=a_1+(a_1+d)+(a_1+2d)
=3a_1+3d
18=3a_1+3d
a_1+d=6\qquad\qquad (1)
a_2=6
a_3=a_1+2d
=a_1+d+d
=6+d
In summary, the terms of the arithmetic sequence are
a_1=a_1
a_2=6
a_3=6+d\qquad\qquad (2)
The terms of the geometric sequence are
a_1+1=a_1+1
a_2=6
a_3+2=6+d+2
=8+d
By definition of geometric sequence
\dfrac{6}{a_1+1}=\dfrac{8+d}{6}
36=(a_1+1)(8+d)
The factors of 36 are 1,2,3,4,6,9,12,18, and 36 from which we form the following sequences

Of these sequences only 3,6,12 and 12,6,3 are geometric sequences. Thus, all possible values for a_1 are 2 and 11. Their sum equals 2+11=13.

Answer: 13
Alternative solution
a_1=a_1
a_2=6
a_3=6+d from Eq. (2)
a_1+a_3=a_1+6+d
=(a_1+d)+6
=6+6 from Eq. (1)
=12\qquad\qquad\qquad (3)
By definition of geometric sequence
\dfrac{6}{a_1+1}=\dfrac{a_3+2}{6}
Substitute the value of a_3 from Eq. (3)
\dfrac{6}{a_1+1}=\dfrac{12-a_1+2}{6}
36=(a_1+1)(14-a_1)
=14a_1-a_1^2+14-a_1
a_1^2-13a_1+22=0
(a_1-2)(a_1-11)=0
a_1=2 or a_1=11

Advertisements

About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s