All Categories MCQs
Topic Notes: All Categories
General Description
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
4221
Find the sum of the first 50 even natural numbers.
Answer:
2550
The sum of the first n even numbers is n(n+1). For n=50, the sum is 50(51) = 2550.
4222
If the lengths of the sides of a right-angled triangle are in AP, the ratio of its sides is:
Answer:
3:4:5
Let the sides be a-d, a, a+d. Since it's a right triangle, the longest side is the hypotenuse. By Pythagorean theorem: (a-d)^2 + a^2 = (a+d)^2. Expanding gives a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2. Simplifying yields a^2 = 4ad. Since length cannot be zero, a = 4d. The sides are 4d-d, 4d, 4d+d, which simplifies to 3d, 4d, 5d. The ratio is 3:4:5.
4223
In an AP, the sum of the first 10 terms is -150 and the sum of the next 10 terms is -550. Find the AP.
Answer:
3, -1, -5, ...
S_10 = 10/2 * (2a + 9d) = 5(2a + 9d) = -150 => 2a + 9d = -30. The sum of the next 10 terms is S_20 - S_10 = -550, so S_20 = -550 + (-150) = -700. Using formula for S_20: 20/2 * (2a + 19d) = 10(2a + 19d) = -700 => 2a + 19d = -70. Subtract the two equations: 10d = -40 => d = -4. Substitute d back: 2a + 9(-4) = -30 => 2a - 36 = -30 => 2a = 6 => a = 3. The AP is 3, -1, -5, ...
4224
Find the sum of the series 1*3 + 3*5 + 5*7 + ... to n terms.
Answer:
n(4n^2 + 6n - 1)/3
The nth term is a_n = (2n-1)(2n+1) = 4n^2 - 1. We need to sum this from 1 to n: Sum(4n^2 - 1) = 4*Sum(n^2) - Sum(1) = 4*n(n+1)(2n+1)/6 - n = n [ 2(2n^2+3n+1)/3 - 1 ] = n [ (4n^2+6n+2-3)/3 ] = n(4n^2 + 6n - 1)/3.
4225
The sum of the first n terms of the series 1 + 3 + 6 + 10 + 15 + ... is:
Answer:
n(n+1)(n+2)/6
The terms are triangular numbers, where the nth term is given by n(n+1)/2. We need to find the sum of these terms: Sum[n(n+1)/2] = (1/2)Sum(n^2+n) = (1/2)[n(n+1)(2n+1)/6 + n(n+1)/2] = (1/2)[n(n+1)/2 * ((2n+1)/3 + 1)] = (1/2)[n(n+1)/2 * (2n+4)/3] = n(n+1)(n+2)/6.
4226
If the sum of an infinite GP is 9 and its second term is 2, find the first term.
Answer:
3 or 6
We have S = a/(1-r) = 9, so a = 9(1-r). The second term is ar = 2, so a = 2/r. Equating the two gives 9(1-r) = 2/r, leading to the quadratic 9r - 9r^2 = 2, or 9r^2 - 9r + 2 = 0. Factoring gives (3r-1)(3r-2)=0. So r = 1/3 or r = 2/3. If r = 1/3, a = 2/(1/3) = 6. If r = 2/3, a = 2/(2/3) = 3. Both 3 and 6 are possible first terms.
4227
If 1/x, 1/y, 1/z are in AP, then x, y, z are in:
Answer:
HP
By definition, a sequence is a Harmonic Progression (HP) if the reciprocals of its terms form an Arithmetic Progression (AP). Thus, x, y, z form an HP.
4228
The harmonic mean of 2 and 8 is:
Answer:
3.2
The formula for Harmonic Mean is 2ab / (a+b). For a=2 and b=8, HM = (2 * 2 * 8) / (2 + 8) = 32 / 10 = 3.2.
4229
Which of the following series does not have a finite sum?
Answer:
2 + 4 + 8 + 16 + ...
An infinite geometric series has a finite sum only if the absolute value of its common ratio is strictly less than 1 (|r| < 1). The series 2 + 4 + 8 + 16 + ... has a common ratio of 2, so its sum diverges to infinity.
4230
If the nth term of a progression is given by a_n = 3n - 5, find the sum of the first 10 terms.
Answer:
115
Since a_n is linear, it is an AP. The first term a_1 = 3(1)-5 = -2. The 10th term a_10 = 3(10)-5 = 25. The sum of the first 10 terms is S_10 = (n/2)(a_1 + a_10) = (10/2)(-2 + 25) = 5(23) = 115.