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
121
What is the sum of the series 1 - 3 + 5 - 7 + ... up to n terms if n is an even number?
Answer:
-n
Since n is even, we can group the terms into pairs: (1 - 3) + (5 - 7) + (9 - 11) + ... There are n/2 pairs. Each pair evaluates to -2. The total sum is (-2) * (n/2) = -n.
122
Find the sum of all two-digit numbers which, when divided by 4, leave a remainder of 1.
Answer:
1210
The numbers are of the form 4k+1. The smallest two-digit number is for k=3 (13) and the largest is for k=24 (97). The AP is 13, 17, 21, ..., 97. Number of terms n = (97 - 13)/4 + 1 = 84/4 + 1 = 21 + 1 = 22. Sum = (n/2)(a + l) = (22/2)(13 + 97) = 11(110) = 1210.
123
If S_n denotes the sum of the first n terms of an AP, and S_{2n} = 3S_n, find the ratio S_{3n} / S_n.
Answer:
6
S_n = (n/2)[2a + (n-1)d]. Given S_{2n} = 3S_n, we have (2n/2)[2a + (2n-1)d] = 3(n/2)[2a + (n-1)d]. This simplifies to 2[2a + (2n-1)d] = 3[2a + (n-1)d]. Expanding: 4a + 4nd - 2d = 6a + 3nd - 3d. Rearranging gives 2a = nd + d = (n+1)d. Now find the ratio S_{3n} / S_n = { (3n/2)[2a + (3n-1)d] } / { (n/2)[2a + (n-1)d] } = 3[2a + (3n-1)d] / [2a + (n-1)d]. Substitute 2a = (n+1)d: = 3[(n+1)d + (3n-1)d] / [(n+1)d + (n-1)d] = 3[4nd] / [2nd] = 12nd / 2nd = 6.
124
Find the sum of the series: 1/2 + 3/4 + 7/8 + 15/16 + ... to n terms.
Answer:
n - 1 + 2^(-n)
The general term is a_k = (2^k - 1) / 2^k = 1 - 1/2^k. The sum to n terms is S_n = Sum(1 - 1/2^k) = Sum(1) - Sum(1/2^k). Sum(1) = n. Sum(1/2^k) is a GP sum: (1/2)[1 - (1/2)^n] / (1 - 1/2) = 1 - (1/2)^n. So, S_n = n - (1 - (1/2)^n) = n - 1 + 2^(-n).
125
The sum of three numbers in an AP is 27 and the sum of their squares is 293. Find the numbers.
Answer:
4, 9, 14
Let numbers be a-d, a, a+d. Sum = 3a = 27, so a = 9. Sum of squares = (9-d)^2 + 9^2 + (9+d)^2 = 293. This gives (81 - 18d + d^2) + 81 + (81 + 18d + d^2) = 293. 243 + 2d^2 = 293 => 2d^2 = 50 => d^2 = 25 => d = 5 (or -5). The numbers are 4, 9, 14.
126
Determine the value of x for which x, x+3, x+9 are in GP.
Answer:
3
For terms to be in GP, the middle term squared equals the product of the other two: (x+3)^2 = x(x+9). Expand: x^2 + 6x + 9 = x^2 + 9x. Subtract x^2 from both sides: 6x + 9 = 9x. Subtract 6x: 9 = 3x, so x = 3.
127
The 3rd term of a GP is 4. The product of its first 5 terms is:
Answer:
4^5
Let the terms be a/r^2, a/r, a, ar, ar^2. The 3rd term is 'a', so a = 4. The product of these 5 terms is (a/r^2) * (a/r) * a * (ar) * (ar^2) = a^5. Since a = 4, the product is 4^5.
128
In a GP, the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.
Answer:
2
We know l = ar^(n-1) and S_n = (lr - a) / (r - 1). Given a=7, l=448, S_n=889. Substituting these values: 889 = (448r - 7) / (r - 1). 889(r - 1) = 448r - 7. 889r - 889 = 448r - 7. 441r = 882, which gives r = 2.
129
Find the sum of the series 1^2 - 2^2 + 3^2 - 4^2 + ... to 2n terms.
Answer:
-n(2n+1)
Group terms in pairs: (1^2 - 2^2) + (3^2 - 4^2) + ... = (1-2)(1+2) + (3-4)(3+4) + ... = -1(3) + -1(7) + -1(11) + ... = -(3 + 7 + 11 + ... to n terms). This is an AP with a=3, d=4. Sum = - (n/2)[2(3) + (n-1)4] = - (n/2)[6 + 4n - 4] = - (n/2)[4n + 2] = -n(2n+1).
130
Find the sum of the series 2^2 + 4^2 + 6^2 + ... + (2n)^2.
Answer:
2n(n+1)(2n+1)/3
Factor out 2^2 from each term: 2^2(1^2 + 2^2 + 3^2 + ... + n^2) = 4 * [n(n+1)(2n+1)/6] = 2n(n+1)(2n+1)/3.