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.

Answer 1

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*}