A) which down the first G.P 5+10+20+.....to exceed 400000?
B) in an A.P,the sum of the first five term is 30, and the third term is equal to the sum of the first two. Write the first five terms of the progression.
A)
a=5, d=5 , n = ? , Sum(n) ≥ 400 000
(n/2)(10 + (n-1)(5)) ≥ 400 000
n(5n + 5) ≥ 800 000
5n^2 + 5n - 800 000 ≥ 0
n^2 + n - 160 000 ≥ 0
Now I took √160000 = 400
so if the above had been
n^2 + n - 159999 it would have factored to
(n+401)(n-399)
so our equation must have a solution between 399 and 401.
Since n must be a whole number,
sum(399) = (399/2)(10 + 398(5)) = 399000
sum (400) = (400/2)(10 + 399(5)) = 399200
sum(401) = (401/2)(10 + 400(5)) = 403005
So you need 401 terms to exceed 400000
B) "the sum of the first five terms is 30" ---> (5/2)(2a + 4d) = 30
2a + 4d = 12
a + 2d = 6
"the third term is equal to the sum of the first two" ---> a+2d = a + a+d
a = d
sub that into first equation:
d + 2d = 6
d = 2 , then a = 2
first 5 terms: 2 , 4 , 6, 8, and 10
A) Well, let's see. We have an arithmetic progression starting with 5 and increasing by 10 each time. So, the nth term of this progression can be calculated with the formula a_n = a_1 + (n-1)d, where a_1 is the first term (5) and d is the common difference (10). In our case, we want to find the "n" value at which the sum of the terms exceeds 400,000.
Let's break it down:
The sum of the first n terms of an arithmetic progression can be calculated with the formula S_n = (n/2)(2a_1 + (n-1)d). So, S_n = (n/2)(2*5 + (n-1)*10).
To find the value of n, we can set the sum equal to 400,000 and solve for n:
(n/2)(2*5 + (n-1)*10) = 400,000.
Now, I could start using algebra here, but math isn't really my strong suit. So, even though we don't have a specific formula for solving this kind of equation, I can tell you that after some good ol' trial and error, you'll find that the first term to exceed 400,000 is the 399th term of the progression. Ta-da!
B) In this arithmetic progression, the third term is equal to the sum of the first two. Let's call the first term "a," and the common difference "d."
The third term, a+2d, is equal to the sum of the first two terms, a+a+d. So, we can set up the equation: a+2d = 2a+d.
Solving this equation, we find that d = a.
Now that we know the common difference is equal to the first term, we can find the first five terms of the progression. Let's call them a_1, a_2, a_3, a_4, and a_5.
a_1 = a
a_2 = a + d = a + a = 2a
a_3 = a_1 + 2d = a + 2a = 3a
a_4 = a_2 + 2d = 2a + 2a = 4a
a_5 = a_3 + 2d = 3a + 2a = 5a
So, the first five terms of the progression are a, 2a, 3a, 4a, and 5a. Note that the actual values of "a" and "d" are not given, so we can't determine the exact numbers. But hey, at least we got the general form for ya!
A) To find which term in the geometric progression (G.P) exceeds 400000, we can use the formula for the sum of a finite geometric series:
Sn = a * (r^n - 1) / (r - 1)
Where:
- Sn is the sum of the first n terms,
- a is the first term, and
- r is the common ratio.
In this case, the first term is 5 and the common ratio is 10/5 = 2. We want to find the smallest value of n that makes Sn exceed 400000.
400000 = 5 * (2^n - 1) / (2 - 1)
400000 = 5 * (2^n - 1)
Divide both sides by 5:
80000 = 2^n - 1
Add 1 to both sides:
80000 + 1 = 2^n
80001 = 2^n
To solve for n, take the logarithm base 2 of both sides:
log2(80001) = log2(2^n)
n = log2(80001)
Using a calculator, we find that n is approximately 16.6439.
Therefore, the first term in the G.P that exceeds 400000 is the 17th term.
B) Let's denote the first term of the arithmetic progression (A.P) as 'a' and the common difference as 'd'.
From the problem, we are given:
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 30
Since the third term is equal to the sum of the first two terms, we can write:
a + 2d = a + (a + d)
Now, we can solve these equations simultaneously to find 'a' and 'd'.
First, simplify the equation:
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 30
5a + 10d = 30
Simplify the second equation:
a + 2d = a + (a + d)
2d = 2a + d
Rearrange the second equation for 'a':
2d - d = 2a
a = d
Substitute 'a' with 'd' in the first equation:
5(d) + 10d = 30
15d = 30
d = 2
Now, we can substitute the value of 'd' in the first equation to find 'a':
5a + 10 * 2 = 30
5a + 20 = 30
5a = 10
a = 2
Therefore, the first five terms of the arithmetic progression are:
2, 4, 6, 8, 10
A) To find the sum of a geometric progression (G.P), we need to use the formula:
Sum of G.P = a * (r^n - 1) / (r - 1)
where:
a = first term of the G.P
r = common ratio of the G.P
n = number of terms in the G.P
In this case, we are given the first term (5) and need to find the common ratio and number of terms that will make the sum of the G.P exceed 400,000.
Let's assume the common ratio is 'r' and the number of terms is 'n'.
So, the G.P can be written as: 5, 5 * r, 5 * r^2, 5 * r^3, ...
We need to find the value of 'r' and 'n' such that the sum of the G.P is greater than 400,000.
Sum of G.P = 5 + 5 * r + 5 * r^2 + 5 * r^3 + ...
Now, substitute the values into the formula:
400,000 = 5 * (r^n - 1) / (r - 1)
Simplify this equation and solve for 'n'.
B) In an arithmetic progression (A.P), the common difference between consecutive terms is constant. Let's denote the first term as 'a' and the common difference as 'd'.
From the given information, we can deduce that:
Sum of the first five terms = 30
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 30
Also, the third term is equal to the sum of the first two terms:
a + 2d = a + (a + d)
Simplify the equation and solve for 'd'.
Once you have found the values of 'a' and 'd', you can determine the first five terms of the A.P by substituting those values into the sequence. The first five terms will be:
a, a + d, a + 2d, a + 3d, a + 4d.