How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
The nth term of a sequence is given 31 - n , find the sum of the first terms of the sequence.
\(\frac{13}{9}\)
1
\(\frac{1}{3}\)
\(\frac{1}{9}\)
Correct answer is A
Tn = 31 - n
S3 = 31 - 1 + 31 - 2 + 31 - 3
= 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\)
= \(\frac{13}{9}\)
Sn is the sum of the first n terms of a series given by Sn = n\(^2\) - 1. Find the nth term
4n + 1
4n - 1
2n + 1
2n - 1
Correct answer is D
\(S_{n} = n^{2} - 1\)
\(T_{n} = S_{n} - S_{n - 1}\)
\(S_{n - 1} = (n - 1)^{2} - 1\)
= \(n^{2} - 2n + 1 - 1\)
= \(n^{2} - 2n\)
\(S_{n} - S_{n - 1} = (n^{2} - 1) - (n^{2} - 2n)\)
= \(2n - 1\)
Find the range of values of x which satisfies the inequality 12x2 < x + 1
-\(\frac{1}{4}\) < x < \(\frac{1}{3}\)
\(\frac{1}{4}\) < x < -\(\frac{1}{3}\)
-\(\frac{1}{3}\) < x < \(\frac{1}{4}\)
-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)
Correct answer is A
12x2 < x + 1
12 - x - 1 < 0
12x2 - 4x + 3x - 1 < 0
4x(3x - 1) + (3x - 1) < 0
Case 1 (+, -)
4x + 1 > 0, 3x - 1 < 0
x > -\(\frac{1}{4}\)x < \(\frac{1}{3}\)
Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0
-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)
Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1
x < - \(\frac{3}{4}\)
x > - \(\frac{4}{3}\)
x > - \(\frac{3}{4}\)
x > - 12
Correct answer is C
\(\frac{f(x)}{g(x)}\) < 1
∴ \(\frac{2x +4}{6x + 7}\) < 1
= 2x + 4 < 6x + 7
= 6x + 7 > 2x + 4
= 6x - 2x > 4 - 7
= 4x > -3
∴ x > -\(\frac{3}{4}\)
3
2
1
-1
Correct answer is A
5 + 2r = k(r + 1) + L(r - 2)
but r - 2 = 0 and r = 2
9 = 3k
k = 3