The first and the last term of an arithmetic progression are in the ratio of 3:4. The sum of all terms is 210 and there are 6 terms in total. Commute common difference.
Let's denote the first term as a and the common difference as d.
We can write the first term as a and the last term as a + 5d (since there are 6 terms in total).
Given that the first term is 3/4 times the last term, we have the equation:
a = (3/4)(a + 5d)
Expanding the equation, we get:
4a = 3a + 15d
Simplifying, we have:
a = 15d
Since the sum of all terms is 210, we can express this as:
(6/2)(a + (a + 5d)) = 210
Simplifying further, we have:
3(a + (a + 5d)) = 210
3(2a + 5d) = 210
6a + 15d = 210
Since we know a = 15d from the previous equation, we can substitute this into the equation:
6(15d) + 15d = 210
90d + 15d = 210
105d = 210
Dividing both sides by 105, we get:
d = 2
Therefore, the common difference is 2.
To find the common difference, we need to use the formulas for the first term (a1), the last term (an), and the sum of an arithmetic progression (Sn).
Given ratio: The first term (a1) and the last term (an) are in the ratio of 3:4. Let's assign a variable to the first term: a1 = 3x. Then, the last term will be: an = 4x.
Number of terms: There are 6 terms in total. We can represent this using the formula: n = 6.
Sum of terms: The sum of the arithmetic progression is given as 210. We can express this using the formula: Sn = (n/2)(a1 + an).
Now, let's substitute the known values and solve the equations:
First, substitute the values for a1 and an:
a1 = 3x
an = 4x
Next, substitute for a1 and an in the formula for the sum of terms:
Sn = (n/2)(a1 + an)
210 = (6/2)(3x + 4x)
210 = 3(7x)
210 = 21x
x = 210/21
x = 10
Finally, substitute the value of x back into the equation for the common difference:
d = an - a1
d = 4x - 3x
d = 4(10) - 3(10)
d = 40 - 30
d = 10
Therefore, the common difference (d) is 10.
Let's start by using the information given to find the first and last terms of the arithmetic progression.
Let's assume the first term is "a" and the common difference is "d".
We are given that the ratio of the first and last terms is 3:4. So, we can write:
last term / first term = 4/3
Using the formula for the nth term of an arithmetic progression, we can express the first term and the last term in terms of the common difference:
first term = a
last term = a + (n-1)d
Since there are 6 terms in total (n = 6), we have:
last term = a + (6-1)d = a + 5d
Substituting these expressions into the ratio equation, we have:
(a+5d) / a = 4/3
Now, let's use the second piece of information given: the sum of all terms is 210.
The formula for the sum of an arithmetic progression is:
sum = (n/2) * (first term + last term)
Substituting the given values, we have:
210 = (6/2) * (a + (a+5d))
Simplifying, we get:
210 = 3*(2a + 5d)
Now, we have two equations:
(a+5d) / a = 4/3
210 = 3*(2a + 5d)
Let's solve this system of equations step by step.
From the first equation, we can cross multiply and simplify:
3(a+5d) = 4a
3a + 15d = 4a
15d = 4a - 3a
15d = a
Now we can substitute this value into the second equation:
210 = 3*(2a + 5d)
210 = 3*(2a + 5(a/15))
210 = 3*(2a + a/3)
210 = 3*((6a + a)/3)
210 = 3*(7a/3)
210 = 7a
a = 210/7
a = 30
Now, we can find the common difference "d" by substituting the value of "a" back into the first equation:
15d = a
15d = 30
d = 30/15
d = 2
Therefore, the common difference is 2.