# Math Help - Arithmetic progression

1. ## Arithmetic progression

The 5th term of an AP is -3. The 6th term is equal to the 1st term multiplied by (-1).
Find the first term and the common difference.
Find the sum of the first 25 terms of the AP.

can anyone help me get started on this?

many thanks

2. Originally Posted by decoy808
The 5th term of an AP is -3. The 6th term is equal to the 1st term multiplied by (-1).
Find the first term and the common difference.
Find the sum of the first 25 terms of the AP.

can anyone help me get started on this?

many thanks
Formulae:

nth term: $U_n = a+(n-1)d$

Sum of n terms: $S_n = \frac{n}{2}[2a+(n-1)d]$

We know:

$U_5 = a + 4d = -3$

$U_6 = a + 5d = -a$

Solve the simultaneous equations for a and d then sub into the above

3. Originally Posted by decoy808
The 5th term of an AP is -3. The 6th term is equal to the 1st term multiplied by (-1).
Find the first term and the common difference.
Find the sum of the first 25 terms of the AP.

can anyone help me get started on this?

many thanks
Let a= 1st term
Let d= difference common between terms.

The sequence is..

$T_1,\ \ T_2,\ \ \ \ \ T_3,\ \ \ \ \ T_4,\ \ \ \ \ \ T_5,\ \ \ \ \ \ \ T_6,....$
a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d),....

The 5th term (a+4d) is -3
The 6th term (a+5d) is -a

Then the difference between these is (a+5d)-(a+4d)=5d-4d=d=-a-(-3)=3-a
Also, a+5d = -a, so 2a+5d = 0, so 2a = -5d = -5(3-a) = 5a-15.

2a=5a-15
15=5a-2a=3a

Solving for "a" allows "d" to be determined.

The sum of an arithmetic sequence is $(number\ of\ terms)(average\ value)$

The average is $\frac{1st\ term+last\ term}{2}$

Hence $T_{25}=a+24d$

The first term is "a".

$\frac{(T_{25}+a)25}{2}=S_{25}$