The 10th, 4th and 1st terms of an AP are the three consecutive terms of a GP. find the common ratio of the GP and the sum of the 1st six terms, taking the 1st term of the AP to be 4?
3 answers
I'm sure Kwara State Polytechni, IIorin is a very interesting School Subject, but we have no tutors skilled in that School Subject. We do have Math tutors, though.
the three terms as an AP would be
a+9d, a+3d, and a
if they also form a GP, then
(a+3d)/(a+9d) = a/(a+3)
a^2 + 6ad + 9d^2 = a^2 + 9ad
9d^2 = 3ad, but a=4
9d^2 = 12d
9d^2 - 12d = 0
3d(3d - 4) = 0
d = 0, or d = 4/3 , we reject d = 0,
so our three AP terms are:
4+9(4/3) , 4 + 3(4/3) and 4
or
16, 8 and 4 , sure enough they also form a GP with
r = 1/2
Do you want the sum of the first 6 terms of the AP or the GP ?
Either way, I am sure you can continue from this point.
a+9d, a+3d, and a
if they also form a GP, then
(a+3d)/(a+9d) = a/(a+3)
a^2 + 6ad + 9d^2 = a^2 + 9ad
9d^2 = 3ad, but a=4
9d^2 = 12d
9d^2 - 12d = 0
3d(3d - 4) = 0
d = 0, or d = 4/3 , we reject d = 0,
so our three AP terms are:
4+9(4/3) , 4 + 3(4/3) and 4
or
16, 8 and 4 , sure enough they also form a GP with
r = 1/2
Do you want the sum of the first 6 terms of the AP or the GP ?
Either way, I am sure you can continue from this point.
sum of six terms in the AP is 31.5
1 answer