In A.P
a = initial term
d = common difference
n-th term:
an = a + ( n - 1 ) d
a1 = a
a2 = a + d
a3 = a + 2 d
In this case
a1 = x
a2 = 3 x + 1
a3 = 7 x - 4
so:
a1 = x
a2 = a1 + d
3 x + 1 = x + d
a3 = a1 + 2 d
7 x - 4 = x + 2 d
Now you must solve system:
3 x + 1 = x + d
7 x - 4 = x + 2 d
First equation.
3 x + 1 = x + d
Subtract x to both sides.
2 x + 1 = d
d = 2 x + 1
Second equation.
7 x - 4 = x + 2 d
Subtract x to both sides.
6 x - 4 = 2 d
2 d = 6 x - 4
Divide both sides by 2.
d = 3 x - 2
Now:
d = d
2 x + 1 = 3 x - 2
Subtract 2x to both sides.
1 = x - 2
Add 2 to both sides.
3 = x
x = 3
d = 2 x + 1
d = 2 • 3 + 1 = 6 + 1
d = 7
Or
d = 3 x - 2
d = 3 • 3 - 2 = 9 - 2
d = 7
1)
a = a1 = x
a = 3
2)
a10 = a1 + 9 d
a10 = 3 + 9 ∙ 7 = 3 + 63 = 66
Your A.P.
3 , 10 , 17 , 24 , 31 , 38 , 45 , 52 , 59 , 66 ...
The first three terms of an A.P are x, (3x+1), (7x-4) .find (1) the value of a
(2) 10th term
1 answer