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
201
If x, 2x+2, 3x+3 are in GP, then what is the 4th term?
Answer:
-13.5
For terms to be in GP, the square of the middle term equals the product of the first and third. (2x+2)^2 = x(3x+3). Expanding gives 4x^2 + 8x + 4 = 3x^2 + 3x. Rearranging yields x^2 + 5x + 4 = 0. Factoring gives (x+4)(x+1) = 0. If x=-1, terms are -1, 0, 0, which is a trivial GP with r=0. If x=-4, terms are -4, -6, -9. Here r = -6/-4 = 1.5. The 4th term is -9 * 1.5 = -13.5.
202
The sum of an infinite geometric series is 15, and its first term is 5. What is the common ratio?
Answer:
2/3
The formula for the sum of an infinite GP is S = a / (1 - r). We are given S = 15 and a = 5. So, 15 = 5 / (1 - r). This means 1 - r = 5 / 15 = 1/3. Therefore, r = 1 - 1/3 = 2/3.
203
What is the 5th term of the geometric sequence 3, -6, 12, ...?
Answer:
48
This is a GP where the first term a = 3 and the common ratio r = -6 / 3 = -2. The formula is a_n = a * r^(n-1). For the 5th term, a_5 = 3 * (-2)^(5-1) = 3 * (-2)^4 = 3 * 16 = 48.
204
Find the sum of all two-digit odd numbers.
Answer:
2475
The two-digit odd numbers are 11, 13, 15, ..., 99. This is an AP where a = 11, d = 2, and the last term l = 99. First, find n: 99 = 11 + (n-1)2 => 88 = (n-1)2 => n-1 = 44 => n = 45. The sum S = (n/2)(a + l) = (45/2)(11 + 99) = (45/2)(110) = 45 * 55 = 2475.
205
If the sum of n terms of a sequence is S_n = 2n^2 + 3n, what is the nth term?
Answer:
4n + 1
The nth term a_n is given by S_n - S_{n-1}. So, a_n = (2n^2 + 3n) - [2(n-1)^2 + 3(n-1)]. Expanding this: a_n = 2n^2 + 3n - [2(n^2 - 2n + 1) + 3n - 3] = 2n^2 + 3n - [2n^2 - 4n + 2 + 3n - 3] = 2n^2 + 3n - [2n^2 - n - 1]. This simplifies to 3n - (-n) - (-1) = 4n + 1.
206
A sequence is generated by the formula a_n = 5n - 2. What kind of sequence is this?
Answer:
Arithmetic Progression
The formula for the nth term is linear with respect to n (of the form an + b). The difference between consecutive terms a_n - a_{n-1} = (5n - 2) - (5(n-1) - 2) = 5, which is a constant. Therefore, it is an Arithmetic Progression with a common difference of 5.
207
The sum of the first n odd natural numbers is:
Answer:
n^2
The first n odd natural numbers are 1, 3, 5, ..., 2n-1. This is an AP with a = 1 and d = 2. Sum = (n/2) * [2(1) + (n-1)2] = (n/2) * [2 + 2n - 2] = (n/2) * [2n] = n^2.
208
The sum of the first n even natural numbers is:
Answer:
n(n+1)
The first n even natural numbers are 2, 4, 6, ..., 2n. This is an AP with a = 2 and d = 2. Sum = (n/2) * [2(2) + (n-1)2] = (n/2) * [4 + 2n - 2] = (n/2) * [2n + 2] = n(n + 1).
209
The 6th term of an AP is zero. What is the relation between its 33rd term and 15th term?
Answer:
a_33 = 3 * a_15
Given a_6 = a + 5d = 0, which means a = -5d. Now, find a_33 and a_15. a_15 = a + 14d = -5d + 14d = 9d. a_33 = a + 32d = -5d + 32d = 27d. Comparing them, we see that 27d is exactly 3 times 9d, so a_33 = 3 * a_15.
210
Find the number of terms in the AP: 7, 13, 19, ..., 205.
Answer:
34
Here, a = 7, d = 6, and the nth term a_n = 205. Using a_n = a + (n-1)d, we have 205 = 7 + (n-1)6. Subtracting 7 gives 198 = (n-1)6. Dividing by 6 yields 33 = n - 1, so n = 34. There are 34 terms.