Find the 10th term of a nonconstant arithmetic sequence whose 1st term is 3 with the 1st, 4th, and 13th term forming a geometric sequence.
I've tried doing it differently everytime, but I always come to a dead end..
Don't answer this anymore.... I got it
In arithmetic sequence :
an = a1 + ( n - 1 ) d
Where:
a1 is the first term of the sequence
d is the common difference
n is the number of the term
a1 = 3
a4 = 3 + ( 3 - 1 ) d = 3 + 2 d
a13 = 3 + ( 13 - 1 ) d = 3 + 12 d
1st, 4th, and 13th term forming a geometric sequence, so :
a1 = 3
a4 = a1 * r = 3 r
a13 = a1 * r ^ 2 = 3 r ^ 2
r is the common ratio
a4 = a4
3 + 2 d = 3 r
a13 = a13
3 + 12 d = 3 r ^ 2
Now you must solve system of two equations :
3 + 2 d = 3 r
3 + 12 d = 3 r ^ 2
3 + 2 d = 3 r Subtract 3 to both sides
3 + 2 d - 3 = 3 r - 3
2 d = 3 r - 3
2 d = 3 ( r - 1 ) Divide both sides by 2
d = 3 ( r - 1 ) / 2
3 + 12 d = 3 r ^ 2
3 + 12 * 3 ( r - 1 ) / 2 = 3 r ^ 2
3 + 36 ( r - 1 ) / 2 = 3 r ^ 2
3 + 18 ( r - 1 ) = 3 r ^ 2 Divide both sides by 3
1 + 6 ( r - 1 ) = r ^ 2
1 + 6 r - 6 = r ^ 2
6 r - 5 = r ^ 2 Subtract r ^ 2 to both sides
6 r - 5 - r ^ 2 = r ^ 2 - r ^ 2
- r ^ 2 + 6 r - 5 = 0 Multiply both sides by - 1
r ^ 2 - 6 r + 5 = 0
Solutions are :
r = 1 and r = 5
For r = 1
d = 3 ( r - 1 ) / 2 = 3 ( 1 - 1 ) / 2 = 3 * 0 / 2 = 0 / 2 = 0
For r = 5
d = 3 ( r - 1 ) / 2 = 3 ( 5 - 1 ) / 2 = 3 * 4 / 2 = 12 / 2 = 6
You have a nonconstant arithmetic sequence so solution r = 1 and d = 0 you must ignore becouse for d = 0 :
an = a1 + ( n - 1 ) d
an = a1 + ( n - 1 ) * 0
an = a1
This mean for d = 0 all members of sequence are equal a1.
That is a constant arithmetic sequence .
So :
r = 5
d = 6
an = a1 + ( n - 1 ) d
a13 = 3 + ( 13 - 1 ) * 6
a13 = 3 + 12 * 6 = 3 + 72 = 75
By the way a geometric sequence of 1st, 4th, and 13th therm :
a1 = 3
a4 = a1 * r = 3 * 5 = 15
a13 = a1 * r ^ 2 = 3 * 5 ^ 2 = 3 * 25 = 75
So a13 = 75
an = a1 + ( n - 1 ) d
a10 = 3 + ( 10 - 1 ) * 6
a10 = 3 + 9 * 6 = 3 + 54 = 57
"a4 = 3 + ( 3 - 1 ) d = 3 + 2 d "
Should be
a4 = 3 + ( 4 - 1 ) d = 3 + 3 d
To find the 10th term of the arithmetic sequence, we first need to find the common difference (d).
Since the 1st, 4th, and 13th terms form a geometric sequence, we can use this information to find the common ratio (r) of the geometric sequence.
Let's denote the 1st term of the geometric sequence as a, and the common ratio as r.
From the information given, we know:
1st term = a
4th term = ar^3
13th term = ar^12
Now we can set up two equations to find the values of a and r.
First, using the 1st and 4th terms:
ar^3 = 3
Secondly, using the 1st and 13th terms:
ar^12 = 3
Now, divide the second equation by the first equation to eliminate a:
(ar^12) / (ar^3) = 3 / 3
Simplifying the equation, we get:
r^9 = 1
Since r^9 = 1, r = 1^(1/9) = 1
Now that we have the value of r, we can find the common difference (d) of the arithmetic sequence. Using the 1st and 4th terms:
4 - 1 = 3 - d
Simplifying the equation, we get:
3 = d
Now that we know the common difference (d = 3), we can find the 10th term of the arithmetic sequence by using the formula:
nth term = a + (n - 1) * d
Plugging in the values, we get:
10th term = 3 + (10 - 1) * 3
= 3 + 9 * 3
= 3 + 27
= 30
Therefore, the 10th term of the given arithmetic sequence is 30.