The first and the last time of an arithmetic progression are in the ratio of 3:4. The sum of all terms is 210 and their are 6 terms in total compute common difference.
Let's assume the first term of the arithmetic progression is "a" and the common difference is "d".
According to the given information, the first and last terms are in the ratio of 3:4.
So, the first term (a) and the last term (a + 5d, since there are 6 terms in total and the common difference is "d") can be written as:
first term = 3x
last term = 4x
Now we can use the formula for the sum of terms in an arithmetic progression:
Sum = (n/2)(first term + last term)
Substituting the given values into the formula:
210 = (6/2)(3x + 4x)
210 = 3(7x)
210 = 21x
x = 210/21
x = 10
Hence, the common difference (d) is equal to "10".
To find the common difference of an arithmetic progression, we can use the formula:
Common difference (d) = (Last term - First term) / (Number of terms - 1).
In this case, we are given that the first and last terms are in the ratio of 3:4. Let's call the first term "a" and the last term "b". So we have:
a:b = 3:4.
We are also given that the sum of all the terms is 210 and there are 6 terms in total.
Let's break down the information given into equations:
First term (a) = 3k
Last term (b) = 4k
Number of terms (n) = 6
We can also represent the sum of all the terms using the formula:
Sum of terms (S) = (n/2) * (2a + (n-1)d),
where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
Plugging in the values we have:
210 = (6/2) * (2(3k) + (6-1)d).
Simplifying the equation:
210 = 3(6k + 5d).
Dividing both sides of the equation by 3:
70 = 6k + 5d.
Now, we can solve this equation simultaneously with the first equation (a:b = 3:4) to find the values of k and d.
Since a:b = 3:4, we can write:
a/b = 3/4.
Substituting the values of a and b:
(3k) / (4k) = 3/4.
Cross-multiplying:
12k = 12k.
This equation shows that k can be any non-zero value. Therefore, we can choose any convenient value for k. For simplicity, let's choose k = 1.
Substituting k = 1 into the equation 70 = 6k + 5d:
70 = 6(1) + 5d.
Simplifying:
70 = 6 + 5d.
Subtracting 6 from both sides:
64 = 5d.
Dividing both sides by 5:
d = 64/5.
Therefore, the common difference is d = 12.8.
Let the first term of the arithmetic progression be $a$, and let the common difference be $d$.
The first term and last term are in the ratio $3:4$, so the last term can be expressed as $4a$. Since there are $6$ terms in total, the sum of the arithmetic progression is $\frac{6}{2}(a + 4a) = 3(5a) = 15a$. Setting this equal to $210$, we can solve for $a$ as follows: \begin{align*}
15a &= 210 \\
\Rightarrow\qquad a &= \frac{210}{15} = 14.
\end{align*}Thus the first term is $14$.
Since the first and last term differ by $(4-1) = 3$ times the common difference, we have $4a - a = 3d$. Substituting in the value we found for $a$, we get \begin{align*}
4a - a &= 3d \\
\Rightarrow\qquad 4(14) - 14 &= 3d \\
\Rightarrow\qquad d &= \frac{4(14) - 14}{3} \\
&= \frac{56- 14}{3} \\
&= \frac{42}{3} = \boxed{14}.
\end{align*}