find the 20th term of an arithmetic progression whose 6th term is 3 and 14th term is 19
10
If the 14th term of an ap is twice it's 8th term and it's 6th term is -8,find the sum of its 20 term
To find the 20th term of an arithmetic progression, we can use the formula:
\[a_n = a_1 + (n-1)d\]
where \(a_n\) is the \(n\)th term of the arithmetic progression, \(a_1\) is the first term, \(n\) is the position of the term, and \(d\) is the common difference between the terms.
We are given that the 6th term (\(a_6\)) is 3 and the 14th term (\(a_{14}\)) is 19. Let's use this information to find the first term (\(a_1\)) and the common difference (\(d\)).
To find \(d\), we can use the following formula:
\[\begin{aligned}
d &= \frac{{a_n - a_1}}{{n - 1}}
\end{aligned}\]
Using the values we have, we can substitute them into the formula:
\[\begin{aligned}
d &= \frac{{a_{14} - a_1}}{{14 - 1}} \\
d &= \frac{{19 - a_1}}{13}
\end{aligned}\]
Next, we can use the second equation given to find \(a_1\). Since the 6th term (\(a_6\)) is 3, we can substitute these values into the formula:
\[\begin{aligned}
a_6 &= a_1 + (6-1)d \\
3 &= a_1 + 5d \\
\end{aligned}\]
Now we have a system of two equations:
\[\begin{aligned}
d &= \frac{{19 - a_1}}{13} \\
3 &= a_1 + 5d \\
\end{aligned}\]
To solve this system of equations, we can substitute the first equation into the second equation:
\[\begin{aligned}
3 &= a_1 + 5\left(\frac{{19 - a_1}}{13}\right) \\
\end{aligned}\]
Now we can solve for \(a_1\) by simplifying and solving the equation.