Mathematics/General Ability MCQs
Topic Notes: Mathematics/General Ability
MCQs and preparation resources for competitive exams, covering important concepts, past papers, and detailed explanations.
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
111
What is the common difference of an AP whose nth term is given by 5 - 3n?
Answer:
-3
The common difference d is the difference between consecutive terms: a_n - a_{n-1}. So, d = (5 - 3n) - (5 - 3(n-1)) = 5 - 3n - 5 + 3n - 3 = -3. Alternatively, in a linear nth term formula An + B, the coefficient of n is the common difference.
112
Find the sum of the first 20 terms of the sequence 1, -2, 3, -4, 5, -6, ...
Answer:
-10
Group the terms in pairs: (1 - 2) + (3 - 4) + (5 - 6) + ... There are 20 terms, so there are 10 pairs. Each pair sums to -1. Total sum = 10 * (-1) = -10.
113
If the p-th, q-th, and r-th terms of a GP are x, y, and z respectively, what is the value of x^(q-r) * y^(r-p) * z^(p-q)?
Answer:
1
Let the GP have first term A and ratio R. Then x = A*R^(p-1), y = A*R^(q-1), z = A*R^(r-1). Substitute these into the expression: [A*R^(p-1)]^(q-r) * [A*R^(q-1)]^(r-p) * [A*R^(r-1)]^(p-q). The exponent of A is (q-r) + (r-p) + (p-q) = 0. The exponent of R is (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q). Expanding and simplifying this also yields 0. Thus, A^0 * R^0 = 1.
114
Find the sum of the series 1 / (1*2) + 1 / (2*3) + 1 / (3*4) + ... to n terms.
Answer:
n / (n+1)
The nth term is 1 / [n(n+1)], which can be written using partial fractions as 1/n - 1/(n+1). This forms a telescoping series: (1/1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1)). All intermediate terms cancel out, leaving 1 - 1/(n+1) = n / (n+1).
115
If a, b, c are in AP and x, y, z are in GP, then what is the value of x^(b-c) * y^(c-a) * z^(a-b)?
Answer:
1
Let a, b, c be a, a+d, a+2d. Then b-c = -d, c-a = 2d, a-b = -d. Let x, y, z be X, XR, XR^2. The expression becomes X^(-d) * (XR)^(2d) * (XR^2)^(-d) = X^(-d + 2d - d) * R^(2d - 2d) = X^0 * R^0 = 1.
116
The sum of the first p terms of an AP is q, and the sum of the first q terms is p. What is the sum of the first (p+q) terms?
Answer:
-(p+q)
This is a standard AP property. If S_p = q and S_q = p, then S_{p+q} is always -(p+q).
117
If the sum of an infinite geometric series is 8 and its second term is 2, what is the common ratio (assuming it's a decreasing series)?
Answer:
1/2
S = a / (1 - r) = 8 => a = 8(1 - r). The second term is ar = 2 => a = 2/r. Equating the two: 8(1 - r) = 2/r. This gives 8r - 8r^2 = 2, or 4r^2 - 4r + 1 = 0. Factoring gives (2r - 1)^2 = 0, so r = 1/2.
118
What is the 5th term of a harmonic progression if its first two terms are 1/2 and 1/4?
Answer:
1/10
If the HP is 1/2, 1/4, ..., then the corresponding AP is 2, 4, ... The first term of the AP is a = 2, and the common difference is d = 4 - 2 = 2. The 5th term of the AP is a_5 = a + 4d = 2 + 4(2) = 10. The 5th term of the HP is the reciprocal, which is 1/10.
119
Find the sum to n terms of the series: 3 + 33 + 333 + ...
Answer:
(1/27)[10(10^n - 1) - 9n]
Let S = 3 + 33 + 333 + ... = 3(1 + 11 + 111 + ...). Multiply and divide by 9: S = (3/9)(9 + 99 + 999 + ...) = (1/3)[(10-1) + (100-1) + (1000-1) + ...]. This becomes (1/3)[(10 + 100 + ...) - (1 + 1 + ...)]. Sum = (1/3) * [ 10(10^n - 1)/9 - n ] = (1/27) * [ 10(10^n - 1) - 9n ]. This matches option d.
120
In a GP, the sum of the first two terms is -4, and the 5th term is 4 times the 3rd term. If the common ratio is positive, find the first term.
Answer:
-4/3
We are given a + ar = -4 and ar^4 = 4ar^2. Assuming a != 0, divide the second equation by ar^2: r^2 = 4. Since the common ratio is positive, r = 2. Substitute r into the first equation: a(1 + 2) = -4 => 3a = -4 => a = -4/3.