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..
4 answers
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
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
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
Should be
a4 = 3 + ( 4 - 1 ) d = 3 + 3 d
1 answer