# Arithmetic sequence

• May 1st 2007, 01:35 PM
Dragon
Arithmetic sequence
The sum of the sixth and the ninth terms of an arithmetic sequence is 20 and their product is 64.Find the tenth term if the first term is negative
• May 1st 2007, 05:43 PM
Jhevon
Quote:

Originally Posted by Dragon
The sum of the sixth and the ninth terms of an arithmetic sequence is 20 and their product is 64.Find the tenth term if the first term is negative

let the sixth term be a
let the ninth term be b

then we have:

a + b = 20 ...........(1)
ab = 64 ...............(2)

From (2) we see that a = 64/b, substitute 64/b for a in (1), we get:

64/b + b = 20
=> 64 + b^2 = 20b
=> b^2 - 20b + 64 = 0
=> (b - 16)(b - 4) = 0
=> b = 16 or b = 4

but, a = 64/b
so when b = 16
a = 64/16 = 4

when b = 4
a = 64/4 = 16

so the solution is: a = 16, b = 4 or a = 4, b = 16

the second choice is correct (that is, a = 4, b = 16), since this will cause the first term to be negative (do you see why)?

Now the terms of an arithmetic sequence is given by:

a_n = a_1 + (n -1)d
where a_n is the current term, a_1 is the first term, n is the current number of the term and d is the common difference.

now, a_6 = 4 = a_1 + (6 - 1)d = a_1 + 5d
and, a_9 = 16 = a_1 + (9 - 1)d = a_1 + 8d

so again we have a system of simultaneous equations:

a_1 + 5d = 4 ...........(1)
a_1 + 8d = 16 ..........(2)

=> 3d = 12 .........(2) - (1)
=> d = 12/3 = 4

but a_1 + 5d = 4
=> a_1 + 5(4) = 4
=> a_1 = 4 - 20 = -16

so our sequence is given by:

a_n = -16 + (n - 1)4
=> a_n = -20 + 4n for n = 1,2,3,4...

so the tenth term is:
a_10 = -20 + 4(10) = 20
• May 1st 2007, 05:53 PM
topsquark
Quote:

Originally Posted by Dragon
The sum of the sixth and the ninth terms of an arithmetic sequence is 20 and their product is 64.Find the tenth term if the first term is negative

The terms of an arithmetic sequence are:
a_n = a_1 + (n - 1)*d
where a_1 is the first term of the series and d is the separation between the terms.

So
a_6 + a_9 = 20 ==> 2*a_1 + 13d = 20

a_6 * a_9 = 64 ==> (a_1)^2 + 13a_1 * d + 40d^2 = 64

From the first equation we get that
d = (2/13)(10 - a_1)

Inserting this into the second equation:
(a_1)^2 + 13a_1*(2/13)(10 - a_1) + 40(2/13)^2 *
(10 - a_1)^2 = 64

Eventually this reduces to:
9(a_1)^2 - 180*a_1 - 5184 = 0

Thus a_1 = 36 or a_1 = -16 by your favorite method of solving quadratics.

Since a_1 is required to be negative, thus a_1 = -16.

So
d = (2/13)(10 - (-16)) = 4.

Thus
a_{10} = -16 + 9*4 = 20

-Dan
• May 1st 2007, 06:03 PM
Jhevon
Quote:

Originally Posted by topsquark
The terms of an arithmetic sequence are:
a_n = a_1 + (n - 1)*d
where a_1 is the first term of the series and d is the separation between the terms.

So
a_6 + a_9 = 20 ==> 2*a_1 + 13d = 20

a_6 * a_9 = 64 ==> (a_1)^2 + 13a_1 * d + 40d^2 = 64

From the first equation we get that
d = (2/13)(10 - a_1)

Inserting this into the second equation:
(a_1)^2 + 13a_1*(2/13)(10 - a_1) + 40(2/13)^2 *
(10 - a_1)^2 = 64

Eventually this reduces to:
9(a_1)^2 - 180*a_1 - 5184 = 0

Thus a_1 = 36 or a_1 = -16 by your favorite method of solving quadratics.

Since a_1 is required to be negative, thus a_1 = -16.

So
d = (2/13)(10 - (-16)) = 4.

Thus
a_{10} = -16 + 9*4 = 20

-Dan

that's nice, cut out the middle man of having to find the 6th and 9th terms
• May 2nd 2007, 03:38 AM
topsquark
Quote:

Originally Posted by Jhevon
that's nice, cut out the middle man of having to find the 6th and 9th terms

Yeah, I hate the middle man. ;)

-Dan
• May 8th 2007, 05:08 PM
Rimas
which one is right

a_10 = -20 + 4(10) = 20
a_{10} = -16 + 9*4 = 20
• May 8th 2007, 05:14 PM
Jhevon
Quote:

Originally Posted by Rimas
which one is right

a_10 = -20 + 4(10) = 20
a_{10} = -16 + 9*4 = 20

both are, it's the same formula:

we used a_n = a_1 + (n - 1)d ..........topsquark left it in this form, i expanded the brackets.

a_n = a_1 + (n - 1)d
=> a_n = a_1 + n*d - d
=> a_n = (a_1 - d) + n*d ................this is the form i used