Given that
and
, if
is an arithmetic sequence, then the common difference between successive terms is
.
You then have
So the explicit formula for the
th term is
d) You have a <u>difference of squares</u>:
49y² - 9 = (7y)² - 3²
Recall the identity,
a² - b² = (a - b) (a + b)
So,
49y² - 9 = (7y - 3) (7y + 3)
e) Pull out the common factor 3 from each term:
3x² - 3x - 90 = 3 (x² - x - 30)
Now use the <u>sum-product method</u>. Notice that we can write 30 = 5 • 6, and 5 - 6 = 1, so
3x² - 3x - 90 = 3 (x + 5) (x - 6)
f) Same as in (e), use the <u>sum-product method</u>. Notice that 42 = 7 • 6, and -7 - 6 = -13, so
x² - 13x + 42 = (x - 7) (x - 6)
Solving this inequality:
-6>t-(-13)
solve as an equation;
6>t+13
6-13>t
-7>t
-7 is greater than t, which means that can be any number smaller that t, fir instance -8, -22 etc.
7 is the third one. 8 is the first one. 9 is the second one.
