The first, second and eleventh terms of an arithmetic sequence with a common difference of 4 are the first, third and fifth terms of a geometric sequence. Find the sequence rule for the geometric and arithmetic sequences.
Thank you for any help!!
The first, second and eleventh terms of an arithmetic sequence with a common difference of 4 are the first, third and fifth terms of a geometric sequence. Find the sequence rule for the geometric and arithmetic sequences.
Thank you for any help!!
Let the arithmetic series be $\displaystyle a_n = a_1 + 4(n - 1)$.
since the first terms of each sequence are the same, we know that the geometric sequence is of the form $\displaystyle b_n = a_1r^{n - 1}$ (the same $\displaystyle a_1$)
this is for $\displaystyle n = 1,2,3, \cdots$
hence, from the arithmetic sequence we have
$\displaystyle a_2 = a_1 + 4$
$\displaystyle a_{11} = a_1 + 40$
from the geometric series we know that
$\displaystyle b_3 = a_1r^2$
$\displaystyle b_5 = a_1r^4$
now, know that $\displaystyle a_1 = b_1, ~a_2 = b_3,~ a_{11} = b_5$ you will be able to set up a system of equations to solve for $\displaystyle a_1$, and hence answer the problem.
(i hope you know the formulas for an arithmetic sequence and geometric sequence to know how i got the expressions for $\displaystyle a_n$ and $\displaystyle b_n$
In A.P:1st term=a, 2nd term=a+4,11th term=a+40
In G.P. 1st term=a, 3rd term=ar^2, 5th term=ar^4
Hence, $\displaystyle ar^2=a+4...(1)$
$\displaystyle ar^4=a+40...(2)$
From (1): $\displaystyle r^2=\frac{a+4}{a}$
Into (2): $\displaystyle a(\frac{a+4}{a})^2=a+40$
$\displaystyle a^2+8a+16=a^2+40$
$\displaystyle a=0.5$
If $\displaystyle a=0.5\rightarrow0.5r^2=4.5$
$\displaystyle r=3$
I'm not sure what you mean by sequence rule, but I've calculated the original term and ratio for the G.P, so I'm not what else you need.
A.P.: $\displaystyle 0.5+4k$
G.P.:$\displaystyle 0.5(3)^k$