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.
If a and b of the first term of r+2 terms find second term before reaching r+2
13 answers
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.
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 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.
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.
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.
The first term and last term are 4 and 20 respectively the sum of terms is 252 calculate number of terms and common difference
Let's denote the first term as 'a', the last term as 'l', the number of terms as 'n', and the common difference as 'd'.
Given that the first term (a) is 4 and the last term (l) is 20, we know that the nth term can be expressed as:
l = a + (n-1)d
Substituting the given values:
20 = 4 + (n-1)d
Simplifying the equation:
16 = (n-1)d
Similarly, we can use the sum of an arithmetic series formula to find the sum of the terms:
S = (n/2)(a + l)
Substituting the given values:
252 = (n/2)(4 + 20)
252 = (n/2)(24)
252 = 12n
Dividing both sides of the equation by 12:
21 = n
Now we have two equations:
16 = (n-1)d
21 = n
Substituting n = 21 into the first equation:
16 = (21 - 1)d
16 = 20d
Dividing both sides of the equation by 20:
4/5 = d
Therefore, the number of terms (n) is 21 and the common difference (d) is 4/5.
Given that the first term (a) is 4 and the last term (l) is 20, we know that the nth term can be expressed as:
l = a + (n-1)d
Substituting the given values:
20 = 4 + (n-1)d
Simplifying the equation:
16 = (n-1)d
Similarly, we can use the sum of an arithmetic series formula to find the sum of the terms:
S = (n/2)(a + l)
Substituting the given values:
252 = (n/2)(4 + 20)
252 = (n/2)(24)
252 = 12n
Dividing both sides of the equation by 12:
21 = n
Now we have two equations:
16 = (n-1)d
21 = n
Substituting n = 21 into the first equation:
16 = (21 - 1)d
16 = 20d
Dividing both sides of the equation by 20:
4/5 = d
Therefore, the number of terms (n) is 21 and the common difference (d) is 4/5.
The carpenter wishes to make a ladder with 15 cross pipe the cross pipe to adminish uniformly in length from 67cm at bottom of 32cm at the top calculate length in each of the seventh cross pipe from the bottom
To calculate the length of the seventh cross pipe from the bottom of the ladder, we need to find the common difference (d) between the lengths of the cross pipes.
Given that the length at the bottom is 67 cm and the length at the top is 32 cm, we can use the formula for the nth term of an arithmetic sequence to find the common difference.
The formula for the nth term (Tn) of an arithmetic sequence is:
Tn = a + (n-1)d
Where a is the first term and d is the common difference.
In this case, the first term (a) is 67 cm and the last term (T15) is 32 cm. We are asked to find the length of the seventh cross pipe from the bottom, which corresponds to the term T7.
Using the formula, we can write:
T15 = a + (15-1)d
32 = 67 + 14d
Simplifying the equation:
14d = -35
d = -35/14
d = -2.5
Now that we have found the common difference (d) as -2.5 cm, we can find the length of the seventh cross pipe from the bottom (T7):
T7 = a + (7-1)d
T7 = 67 + 6(-2.5)
T7 = 67 - 15
T7 = 52 cm
Therefore, the length of the seventh cross pipe from the bottom is 52 cm.
Given that the length at the bottom is 67 cm and the length at the top is 32 cm, we can use the formula for the nth term of an arithmetic sequence to find the common difference.
The formula for the nth term (Tn) of an arithmetic sequence is:
Tn = a + (n-1)d
Where a is the first term and d is the common difference.
In this case, the first term (a) is 67 cm and the last term (T15) is 32 cm. We are asked to find the length of the seventh cross pipe from the bottom, which corresponds to the term T7.
Using the formula, we can write:
T15 = a + (15-1)d
32 = 67 + 14d
Simplifying the equation:
14d = -35
d = -35/14
d = -2.5
Now that we have found the common difference (d) as -2.5 cm, we can find the length of the seventh cross pipe from the bottom (T7):
T7 = a + (7-1)d
T7 = 67 + 6(-2.5)
T7 = 67 - 15
T7 = 52 cm
Therefore, the length of the seventh cross pipe from the bottom is 52 cm.