157
187
197
200
Correct answer is B
a = 2, d = 3 and n = 11
To find Sn/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)]
= \(\frac{11}{2}\) [2(2) + (11 - 1) 3]
= \(\frac{11}{2}\)n [4 + 10(3)]
= \(\frac{11}{2}\)(34)
= 11 x 17
= 187
What is the n-th term of the sequence 2, 6, 12, 20...?
4n - 2
2(3n - 1)
n2 + n
n2 + 3n + 2
Correct answer is C
Given that 2, 6, 12, 20...? the nth term = n\(^2\) + n
check: n = 1, u1 = 2
n = 2, u2 = 4 + 2 = 6
n = 3, u3 = 9 + 3 = 12
∴ n = 4, u4 = 16 + 4 = 20
1270
190
18
9
Correct answer is D
3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
a = 3
r = \(\frac{2}{3}\)
s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)
= \(\frac{3}{\frac{1}{3}}\)
= 3 x 3
= 9
Find all values of x satisfying the inequality -11 \(\leq\) 4 - 3x \(\leq\) 28
-5 \(\leq\) x v 8
5 \(\leq\) x \(\leq\) 8
-8 \(\leq\) x \(\leq\) 5
-5 < x \(\leq\) 8
Correct answer is C
To solve -11 \(\leq\) 4 - 3x \(\leq\) 28
-11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28
15 \(\leq\) -3x \(\leq\) 24 = 15 \(\geq\) 3x - 3x \(\geq\) -24
-5 \(\geq\) x, x \(\geq\) -8
i.e. x \(\leq\) 5
∴ -8 \(\leq\) x \(\leq\) 5
Resolve \(\frac{3}{x^2 + x - 2}\) into partial fractions
\(\frac{1}{x - 1} - \frac{1}{x + 2}\)
\(\frac{1}{x + 1} + \frac{1}{x - 2}\)
\(\frac{1}{x + 1} - \frac{1}{x - 2}\)
\(\frac{1}{x - 2} + \frac{1}{x + 2}\)
Correct answer is A
\(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 2)}\)
\(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\)
A(x + 2) + B(x - 1) = 3
when x = 1, 3A = 3 \(\to\) a = 1
when x = -2, -3B = 3 \(\to\) B = -1
= \(\frac{1}{x - 1} - \frac{1}{x + 2}\)