I have a problem in my maths book which says
Find the arithmetic sequence in which $T_8 = 11$ and $T_{10}$ is the additive inverse of $T_{17}$
I don't have a first term of common difference to solve it, so I managed to make two equations to find the first term and common difference from them. Here they are.
first equation $a + 7d = 11$
second equation since $T_{10} + T_{17} = 0$
therefore $a+9d + a+16d = 0 ~~\Rightarrow~~ 2a + 25d = 0$
So what I did is subtracting the first equation from the second one to form this equation with two unknowns $a – 18d = 11$
This is what I came up with and I can't solve the equations, any help?
To solve linear equations with n variables, you need a minimum of n equations. Here, we have 2 variables and 2 equations, so it can be solved.
There are many methods to solve the equations. But the most common method is elimination method.
In this method, you multiply an equation with a constant, such that the coefficient of the equation becomes equal to the coefficient of the other equation. Then you subtract them, and you are left with one variable.
You have the equations:
$$ a + 7d = 11 $$ $$ 2a + 25d = 0 $$
Now multiply the first equation by 2, you get:
$$2a+14d=22$$
Subtract this with the second equation. You get:
$$ 2a+14d - 2a - 25d = 22$$ $$ - 11d = 22 $$
This gives $d= -2 $. Putting this value in first equation, we get: $$a = 11 - 7.(-2) = 25 $$
So we have, $a=25, d=-2$