PPT
pp-1
qp
pp
Correct answer is A
p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)
Q = \(\begin{vmatrix} 0 & 2 & 4 \\ 3 & 1 & 2\\ 0 & 3 & 2 \end{vmatrix}\) = pT
pq = ppT
48
24
-24
-48
Correct answer is B
p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)
PT = \(\begin{vmatrix}0 & 2 & 4 \\ 2 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)
/pT/ = \(\begin{vmatrix}0 & 2 & 4 \\ 3 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)
= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]
= 0 - 12 + 36 = 24
e = 1
e = -1
e = -2
e = 0
Correct answer is B
Identity(e) : a \(\ast\) e = a
m \(\ast\) e = m...(i)
m \(\ast\) e = me + m + e
Because m \(\ast\) e = m
: m = me + m + e
m - m = e(m + 1)
e = \(\frac{0}{m + 1}\)
e = 0
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