1) In the arithmetic progression, the 8th term is twice the 4th term and the 20th term is 40. Find the common difference and the sum of the terms from the 8th to the 20th term inclusive.
2) Find the least number of terms of the AP, 1+3+5...that are required to make a sum exceeding 4000.
(?a=1,d=2)
Any help at all would be great
1)
remember that t(n) = a+(n-1)d
t(8) = a+7d
t(4) = a+3d
it said: the 8th term is twice the 4th term ---> a+7d = 2(a+3d), simplify this
then it said: the 20th term is 40 ---> a+19d = 40
Solve these two equations.
2) YOu should know that
S(n) = n/2[2a + (n-1)d]
so you want n/2[2 + (n-1)(2)] > 4000
solve this
Th 4th term 0f an a p is 13 and the 10th term is 29 find the 16th term
2n
1) Ah, arithmetic progressions, always marching forward in a fixed manner. Let's unravel this puzzle, shall we?
Given that the 8th term is twice the 4th term, we can write the equation as follows:
a + 7d = 2(a + 3d),
where "a" is the first term and "d" is the common difference.
Simplifying the equation, we get:
a + 7d = 2a + 6d.
Rearranging the terms, we find:
a = d.
Now, moving on to the 20th term being 40, we can set up another equation:
a + 19d = 40.
Since we know that a = d, we can substitute it into the equation:
a + 19(a) = 40.
2a = 40.
a = 20.
So, the common difference (d) is 20, and the first term (a) is also 20.
To find the sum of terms from the 8th to the 20th (inclusive), we can use the formula for the sum of an arithmetic progression:
Sn = (n/2)(2a + (n-1)d),
where "Sn" is the sum of the terms, "n" is the number of terms, "a" is the first term, and "d" is the common difference.
Plugging in the values, we get:
S13 = (13/2)(2(20) + (13-1)(20)).
S13 = (13/2)(40 + 240).
S13 = 13(140).
S13 = 1820.
So, the sum of the terms from the 8th to the 20th (inclusive) is 1820.
2) Ah, let's ramble through this arithmetic progression with a spring in our step!
We have the AP 1 + 3 + 5 + ... with a common difference of 2. We want to find the least number of terms required to make a sum exceeding 4000.
Using the formula for the sum of an arithmetic progression, Sn = (n/2)(2a + (n-1)d), we can plug in the values:
4000 < (n/2)(2(1) + (n-1)(2)).
Simplifying, we get:
4000 < (n/2)(2 + 2n - 2).
4000 < (n/2)(2n).
8000 < n^2.
Therefore, we need a value of "n" greater than the square root of 8000, which is approximately 89.44.
Since we're talking about the number of terms, we round up to the nearest whole number, meaning we need at least 90 terms.
1) To find the common difference and the sum of the terms in an arithmetic progression, we need to use the formulas associated with arithmetic progressions.
Let's use the formulas to solve the problem:
First, let's find the common difference (d):
The general formula to find the nth term (Tn) in an arithmetic progression is: Tn = a + (n - 1) * d
where a is the first term, n is the term number, and d is the common difference.
Given that the 8th term is twice the 4th term, we can write the equation: 8th term = 2 * 4th term
Using the general formula, we have:
a + 7d = 2(a + 3d) [a + 7d represents the 8th term and a + 3d represents the 4th term]
Simplifying the equation:
a + 7d = 2a + 6d
7d - 6d = 2a - a
d = a
So, we have found that the common difference (d) is equal to the first term (a).
Next, let's find the value of the first term (a):
We are given that the 20th term is 40. Using the general formula:
T20 = a + (20 - 1) * d
Substituting the values:
40 = a + 19a [since d = a]
Simplifying the equation:
20a = 40
a = 2
Now that we have found the value of a, we can substitute it back into the first equation to find the common difference (d):
d = a = 2
Therefore, the common difference is 2.
To find the sum of the terms from the 8th to the 20th term (inclusive), we can use the formula for the sum of an arithmetic series.
The formula to find the sum of an arithmetic series is: Sn = (n/2)(2a + (n - 1)d)
where Sn is the sum of the first n terms.
Using this formula:
S = (n/2)(2a + (n - 1)d)
Substituting the values:
S = (13/2)(2(2) + (13 - 1)(2))
= (13/2)(4 + 12)
= (13/2)(16)
= 104
Hence, the sum of the terms from the 8th to the 20th term (inclusive) is 104.
2) To find the least number of terms of the arithmetic progression 1, 3, 5,... required to make a sum exceeding 4000, we need to use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is: Sn = (n/2)(2a + (n-1)d), where Sn is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.
Given that the first term (a) is 1 and the common difference (d) is 2, we can substitute these values into the formula and solve for the number of terms (n):
Sn > 4000
Using the formula:
(n/2)(2(1) + (n-1)(2)) > 4000
Simplifying the equation:
(n/2)(2 + 2n - 2) > 4000
(n/2)(2n) > 4000
n^2 > 4000
n > sqrt(4000)
Taking the square root of both sides:
n > 63.2455
Since the number of terms (n) must be a whole number, we can round up to the nearest whole number:
n > 64
Therefore, the least number of terms required to make a sum exceeding 4000 is 64.
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