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
4201
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.
4202
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).
4203
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.
4204
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.
4205
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.
4206
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.
4207
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).
4208
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.
4209
A person pays a loan of Rs. 975 in monthly installments, each installment being Rs. 5 less than the former. If the first installment is Rs. 100, in how many months will the loan be fully paid?
Answer:
15
The installments form an AP with a=100 and d=-5. We need S_n = 975. Using S_n = (n/2)[2a + (n-1)d], we get 975 = (n/2)[200 + (n-1)(-5)]. 1950 = 200n - 5n^2 + 5n. 5n^2 - 205n + 1950 = 0. Divide by 5: n^2 - 41n + 390 = 0. Factors are (n-15)(n-26)=0. So n=15 or 26. However, at n=26, the installment would be 100 + 25(-5) = -25, which is negative (not physically meaningful for a loan payment). The practical answer is 15.
4210
The sum of the infinite series 1 + 2x + 3x^2 + 4x^3 + ... for |x| < 1 is:
Answer:
1/(1-x)^2
This is a standard Arithmetico-Geometric series. The sum can be derived by differentiating the geometric series sum 1 + x + x^2 + ... = 1/(1-x) with respect to x, which gives 1 + 2x + 3x^2 + ... = 1/(1-x)^2.