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