How many terms of the series -9, -6, -3,... must be taken that the sum may be 66?

The given series is an arithmetic progression with a common difference of 3. To find the number of terms needed to obtain a sum of 66, we can use the formula for the sum of an arithmetic series:

=
2
(
2
+
(
−
1
)
)
S=
2
n
​
(2a+(n−1)d)

where:

S is the sum of the series,
n is the number of terms,
a is the first term, and
d is the common difference.
In this case, we have:

S = 66,
a = -9, and
d = 3.
Substituting these values into the formula, we get:

66
=
2
(
2
(
−
9
)
+
(
−
1
)
(
3
)
)
66=
2
n
​
(2(−9)+(n−1)(3))

Simplifying:

66
=
2
(
−
18
+
3
�
−
3
)
66=
2
n
​
(−18+3n−3)

66
=
2
(
3
−
21
)
66=
2
n
​
(3n−21)

Multiplying both sides by 2 to eliminate the fraction:

132
=
(
3
−
21
)
132=n(3n−21)

132
=
3
2
−
21
132=3n
2
−21n

Rearranging the equation:

3
2
−
21
−
132
=
0
3n
2
−21n−132=0

Dividing through by 3:

2
−
7
−
44
=
0
n
2
−7n−44=0

Now, we can solve this quadratic equation for n. Factoring is the simplest method:

(
−
11
)
(
+
4
)
=
0
(n−11)(n+4)=0

So, n - 11 = 0 or n + 4 = 0.

If n - 11 = 0, then n = 11.

If n + 4 = 0, then n = -4.

Since the number of terms must be positive, we disregard the negative value of n. Therefore, the number of terms needed to obtain a sum of 66 is 11.