The sum of the first term and seventh term of an arithmetic progression is 6.If the twentieth term is 56,find the term whose value first exceeds 400.
The terms are in arithmetic progression, so lets call the difference between terms d, and the first term a. Then we are given that the first term and seventh term sum to 6, therefore,
a + (a + 6d) = 6 .
We are then given that the twentieth term is 56, so
a + 19d = 56 .
So now we have two equations in two unknowns, which we (you) can solve to find a and d. From there you should be able to complete the question.
Hope this helps.
$\displaystyle t_n = a + (n - 1)d$.
$\displaystyle t_1 = a + (1 - 1)d$
$\displaystyle t_1 = a$.
$\displaystyle t_7 = a + (7 - 1)d$
$\displaystyle t_7 = a + 6d$.
$\displaystyle t_1 + t_7 = 6$
$\displaystyle a + a + 6d = 6$
$\displaystyle 2a + 6d = 6$
$\displaystyle a + 3d = 3$.
$\displaystyle t_{20} = 56$
$\displaystyle a + (20 - 1)d = 56$
$\displaystyle a + 19d = 56$.
Solving the equations simultaneously:
$\displaystyle a + 3d = 3$ (1)
$\displaystyle a + 19d = 56$ (2)
(2) - (1):
$\displaystyle (a + 19d)-(a + 3d) = 56 - 3$
$\displaystyle 16d = 53$
$\displaystyle d = \frac{53}{16}$.
So $\displaystyle a + 3d = 3$
$\displaystyle a + 3\left(\frac{53}{16}\right) = 3$
$\displaystyle a + \frac{159}{16} = 3$
$\displaystyle a = \frac{48}{16} - \frac{159}{16}$
$\displaystyle a = -\frac{111}{16}$.
So to find the first value that is over 400...
$\displaystyle t_n = a + (n - 1)d$
$\displaystyle 400 = -\frac{111}{16} + \frac{53}{16}(n - 1)$
$\displaystyle \frac{6400}{16} = -\frac{111}{16} + \frac{53}{16}(n - 1)$
$\displaystyle \frac{6511}{16} = \frac{53}{16}(n - 1)$
$\displaystyle \frac{6511}{53} = n - 1$
$\displaystyle \frac{6564}{53} = n$
$\displaystyle n = 123.85$.
So the first term that i going to be greater than $\displaystyle 400$ is $\displaystyle t_{124}$