the 10th,4th and 1st term of an A.P are the three consecutive numbers of a G.P and the sum of the 1st 6 terms ;take 4 as the 1st term
Reiny did this. See the related questions below.
*Let the three terms of the AP be*
10th term = a + 9d
4th term = a + 3d
1st term = a
∴ (a + 9d), (a + 3d), and a
*If they also form a GP, then*
(a + 3d)/(a + 9d) = a/(a + 3d)
*Cross multiply*
(a + 3d)(a + 3d) = a(a + 9d)
a² + 3ad + 3ad + 9d² = a² + 9ad
a² + 6ad + 9d² = a² + 9ad
∴ a² + 6ad + 9d² - a² - 9ad = 0
∴ 9d² - 3ad = 0
∴ 3ad = 9d²
*But, we know that, a = 4*
∴ 3(4)d = 9d²
∴ 12d = 9d²
∴ 9d² - 12d = 0
∴ 3d(3d - 4) = 0
∴ 3d = 0 or 3d - 4 = 0
∴ d = 0/3 or 3d = 4
∴ d = 0, or d = 4/3
*Since common difference cannot be 0*
∴ d = 4/3
*so our three AP terms are:*
∴ (a + 9d), (a + 3d), and a
= 4 + 9(4/3) , 4 + 3(4/3) and 4
= 4 + 3(4) , 4 + 4 , and 4
= 16, 8 and 4 ,
*sure enough they also form a GP with, Common ratio, r = 1/2*
Now, a = 16
*Now, Sum of the first 6th terms of the G.P*
Sn = a[1 - rⁿ]/(1 - r)
S(6) = 16[1 - (1/2)^6]/(1 - 1/2)
S(6) = 16[1 - (1/64)]/(1/2)
S(6) = 16[(64 - 1)/64]/(1/2)
S(6) = (16 × 2)[(63)/64]
S(6) = (32)[(63)/64]
S(6) = 63/2
S(6) = 31.5
DONE !
Force × Distance = ISE = Dumb physics = Dumb ISE
*(I am a learner !!!)*
To solve this problem, we'll use the formulas for both an arithmetic progression (A.P.) and a geometric progression (G.P.) to find the required terms.
Let's start by considering the arithmetic progression with a common difference (d) of "4".
The general formula for an A.P. is given by:
an = a1 + (n-1)d
Given that "4" is the first term (a1), we can calculate the 10th term (a10), the 4th term (a4), and the 1st term (a1) of the A.P.
a10 = a1 + (10-1)d
= 4 + (9)4
= 4 + 36
= 40
a4 = a1 + (4-1)d
= 4 + (3)4
= 4 + 12
= 16
Now, we know that the 10th, 4th, and 1st terms of the A.P. are three consecutive terms of a geometric progression (G.P.). In a G.P., each term is obtained by multiplying the previous term by a common ratio (r).
Let's assume the common ratio is "r".
Now, we need to find the common ratio (r) of the G.P. that satisfies the given conditions.
a1 x r^2 = a4
4 x r^2 = 16
Dividing both sides by 4, we get:
r^2 = 4
Taking the square root of both sides, we have:
r = ±2
Since the problem states that the three numbers form a G.P, we will consider the positive value of "2" for the common ratio.
Now that we have the common ratio, we can find the terms of the G.P.
The general formula for a G.P. is given by:
an = a1 * r^(n-1)
By substituting the values, we can determine the terms:
G.P. 1st term (b1) = 4
G.P. common ratio (r) = 2
G.P. 1st term (b1) = 4
G.P. 2nd term (b2) = 4 * 2^(2-1) = 4 * 2 = 8
G.P. 3rd term (b3) = 4 * 2^(3-1) = 4 * 4 = 16
Therefore, the 10th term of the A.P. is 40, and the corresponding terms in the G.P. are 8, 16, and 32.
To find the sum of the first 6 terms of the A.P., we can use the formula for the sum of an A.P.:
Sum(n) = (n/2)[2a1 + (n-1)d]
In this case, we want to find the sum of the first 6 terms of the A.P.
n = 6
a1 = 4
d = 4
Substituting the values, we get:
Sum(6) = (6/2)[2(4) + (6-1)(4)]
= 3[8 + 5(4)]
= 3[8 + 20]
= 3[28]
= 84
Therefore, the sum of the first 6 terms of the A.P. is 84.