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?
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.
sum of six terms in the AP is 31.5
To find the common ratio (r) of the geometric progression (GP) and the sum of the first six terms, we need to use the given information.
Let's start by finding the common difference (d) of the arithmetic progression (AP). The first term (a) of the AP is given as 4, and we know that the 10th, 4th, and 1st terms of the AP are in geometric progression. Hence, we can write:
10th term (a + 9d), 4th term (a + 3d), and 1st term (a) are in geometric progression.
Using the geometric progression formula, we get:
(a + 3d) / (a) = (a + 9d) / (a + 3d)
Cross-multiply:
(a)(a + 9d) = (a + 3d)(a + 3d)
Expand and simplify:
a² + 9ad = a² + 6ad + 9d²
Simplify further by canceling out the a² term:
9ad = 6ad + 9d²
3ad = 9d²
Divide both sides by d:
3a = 9d
Now substitute the value of a (first term of the AP) which is 4:
3(4) = 9d
12 = 9d
Divide both sides by 9:
d = 12 / 9
d = 4 / 3
Therefore, the common difference (d) of the arithmetic progression is 4/3.
Now, let's find the common ratio (r) of the geometric progression. We know that the 10th, 4th, and 1st terms of the arithmetic progression are also the terms of a geometric progression. Hence, we can write:
(a + 9d) / (a + 3d) = (a + 3d) / (a)
Substitute the values of a (4) and d (4/3):
(4 + 9(4/3)) / (4 + 3(4/3)) = (4 + 3(4/3)) / (4)
Simplify:
(4 + 12/3) / (4 + 4) = (4 + 4) / (4)
(4 + 4) / (4) = (8) / (4) = 2
Therefore, the common ratio (r) of the geometric progression is 2.
Now, let's find the sum of the first six terms (S6) of the arithmetic progression. We can use the formula for the sum of an arithmetic progression:
S6 = (n/2)(2a + (n - 1)d)
Substitute the values of n (6), a (4), and d (4/3):
S6 = (6/2)(2(4) + (6 - 1)(4/3))
Simplify:
S6 = (3)(8 + 5(4/3))
S6 = (3)(8 + 20/3)
S6 = (3)(24/3 + 20/3)
S6 = (3)(44/3)
S6 = 44
Therefore, the sum of the first six terms of the arithmetic progression is 44.
In summary:
- The common ratio (r) of the geometric progression is 2.
- The sum of the first six terms (S6) of the arithmetic progression is 44.
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.