1. ## Arithmetic progression

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.

2. Originally Posted by mastermin346
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.

3. Originally Posted by mastermin346
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.
$t_n = a + (n - 1)d$.

$t_1 = a + (1 - 1)d$

$t_1 = a$.

$t_7 = a + (7 - 1)d$

$t_7 = a + 6d$.

$t_1 + t_7 = 6$

$a + a + 6d = 6$

$2a + 6d = 6$

$a + 3d = 3$.

$t_{20} = 56$

$a + (20 - 1)d = 56$

$a + 19d = 56$.

Solving the equations simultaneously:

$a + 3d = 3$ (1)

$a + 19d = 56$ (2)

(2) - (1):

$(a + 19d)-(a + 3d) = 56 - 3$

$16d = 53$

$d = \frac{53}{16}$.

So $a + 3d = 3$

$a + 3\left(\frac{53}{16}\right) = 3$

$a + \frac{159}{16} = 3$

$a = \frac{48}{16} - \frac{159}{16}$

$a = -\frac{111}{16}$.

So to find the first value that is over 400...

$t_n = a + (n - 1)d$

$400 = -\frac{111}{16} + \frac{53}{16}(n - 1)$

$\frac{6400}{16} = -\frac{111}{16} + \frac{53}{16}(n - 1)$

$\frac{6511}{16} = \frac{53}{16}(n - 1)$

$\frac{6511}{53} = n - 1$

$\frac{6564}{53} = n$

$n = 123.85$.

So the first term that i going to be greater than $400$ is $t_{124}$

4. EDIT: That step is most definitely not wrong

5. Originally Posted by pomp
That step is wrong (but I suppose that at least gives the OP some work to do )
Doesn't $\frac{6400}{16} = 400$?

6. Originally Posted by Prove It
Doesn't $\frac{6400}{16} = 400$?

Woops! haha yes, sorry I went totally mathslexic for a second there.