Thread: 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

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

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

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

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.

But the advantage of yours is that your equations look nicer.

-Dan

which one is right

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

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