1. ## Arithmatic and Geometric Sequences Help Please?

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!!

2. Originally Posted by anonymous_maths
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 $a_n = a_1 + 4(n - 1)$.

since the first terms of each sequence are the same, we know that the geometric sequence is of the form $b_n = a_1r^{n - 1}$ (the same $a_1$)

this is for $n = 1,2,3, \cdots$

hence, from the arithmetic sequence we have

$a_2 = a_1 + 4$

$a_{11} = a_1 + 40$

from the geometric series we know that

$b_3 = a_1r^2$

$b_5 = a_1r^4$

now, know that $a_1 = b_1, ~a_2 = b_3,~ a_{11} = b_5$ you will be able to set up a system of equations to solve for $a_1$, 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 $a_n$ and $b_n$

3. 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, $ar^2=a+4...(1)$
$ar^4=a+40...(2)$

From (1): $r^2=\frac{a+4}{a}$
Into (2): $a(\frac{a+4}{a})^2=a+40$
$a^2+8a+16=a^2+40$
$a=0.5$

If $a=0.5\rightarrow0.5r^2=4.5$
$r=3$

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.: $0.5+4k$
G.P.: $0.5(3)^k$