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
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