If a and b of the first term of r+2 terms find second term before reaching r+2
To find the second term before reaching r+2 terms, we can subtract 1 from r+2, since the second term will be the term right before the r+2 term.
So, the second term before reaching r+2 terms can be represented as (r+2) - 1, which simplifies to r+1.
Find the number of terms of the series 2+6+10+14+18 that will give sum of 800
The given series is an arithmetic progression with a common difference of 4.
The formula to find the sum of an arithmetic series is:
S = (n/2)(2a + (n-1)d)
Where S is the sum of the series, n is the number of terms, a is the first term, and d is the common difference.
In this case, the first term (a) is 2 and the common difference (d) is 4.
We want the sum of the series to be 800. So we can write the equation:
800 = (n/2)(2(2) + (n-1)(4))
Simplifying the equation:
800 = (n/2)(4 + 4(n-1))
800 = (n/2)(4 + 4n - 4)
800 = (n/2)(4n)
800 = 2n^2
400 = n^2
n^2 = 400
n = √400
n = 20
Therefore, there are 20 terms in the series.
20 terms
Apologies for my previous incorrect response. Let me calculate it correctly.
The given series is an arithmetic progression with a first term (a) of 2 and a common difference (d) of 4.
To find the number of terms (n) that will give a sum of 800, we can use the formula for the sum of an arithmetic series:
S = (n/2)(2a + (n-1)d)
Substituting the given values, we have:
800 = (n/2)(2(2) + (n-1)(4))
800 = (n/2)(4 + 4(n-1))
800 = (n/2)(4 + 4n - 4)
800 = (n/2)(4n)
800 = 2n^2
Dividing both sides by 2, we have:
400 = n^2
Taking the square root of both sides gives:
n = √400
n = 20
Therefore, there are 20 terms in the series that will give a sum of 800.
The eleventh term of AP is four times it's second term the sum of first seven terms of same progression is 175 find first term and common difference
Let's denote the first term of the arithmetic progression as 'a', and the common difference as 'd'.
We are told that the 11th term of the AP is four times its second term, so we can write the equation:
a + 10d = 4(a + d)
Expanding the equation:
a + 10d = 4a + 4d
Rearranging the terms:
10d - 4d = 4a - a
6d = 3a
Now, let's consider the sum of the first seven terms of the AP, denoted as 'S7'. The formula to find the sum of the first n terms of an arithmetic progression is:
Sn = (n/2)(2a + (n-1)d)
Plugging in the given values:
175 = (7/2)(2a + (7-1)d)
175 = (7/2)(2a + 6d)
175 = 7a + 21d
Since we have two equations:
6d = 3a (equation 1)
7a + 21d = 175 (equation 2)
We can solve this system of equations to find the values of 'a' and 'd'.
Multiplying equation 1 by 7, we get:
42d = 21a
Substituting this into equation 2:
21a + 21d = 175
Dividing both sides of the equation by 21, we have:
a + d = 175/21
a + d = 25/3
Multiplying equation 1 by 2, we get:
12d = 6a
Substituting this into a + d = 25/3:
6a + 6d = 25
Subtracting this equation from equation 2, we get:
(21a + 21d) - (6a + 6d) = 175 - 25
15a + 15d = 150
Dividing by 15:
a + d = 10
Since both a + d = 25/3 and a + d = 10, we can equate the expressions:
25/3 = 10
This is not true, which means there is no consistent solution to the system of equations. Therefore, there is no specific first term and common difference that satisfy both conditions given in the problem.
An experimental culture has initial population of 50 bacteria the population increased by 80percent determine the time it will take to have population of 1.2 million bacteria
To determine the time it will take for the population to reach 1.2 million bacteria, we need to consider the growth rate.
Given that the population increases by 80%, we know that the population at any time (t) will be 1.8 times the previous population.
Let P(t) represent the population at time t.
We can express the population at time t as: P(t) = P * (1 + r)^t, where P is the initial population and r is the growth rate.
In this case, the initial population (P) is 50 bacteria, and the growth rate (r) is 80%, or 0.80. We want to find the time (t) when the population reaches 1.2 million bacteria.
1.2 million bacteria can also be expressed as 1,200,000. So, the equation becomes:
P(t) = 50 * (1 + 0.80)^t = 1,200,000
Dividing both sides of the equation by 50:
(1 + 0.80)^t = 24,000
Taking the natural logarithm (ln) of both sides of the equation:
ln(1.80)^t = ln(24,000)
Now, we can solve for t by dividing both sides of the equation by ln(1.80):
t = ln(24,000) / ln(1.8)
Using a calculator, we can find that ln(24,000) ≈ 10.086 and ln(1.8) ≈ 0.5878.
So, t = 10.086 / 0.5878 ≈ 17.152
Therefore, it will take approximately 17.152 units of time for the population to reach 1.2 million bacteria.