Find the 10th term of the series 3, 7, 13, 21, 31, ... using the method of differences.
91
101
111
121
Explanation: Differences: 4, 6, 8, 10 (an AP). tⁿ = a + sum of (n-1) differences = 3 + (n-1)/2[2(4) + (n-2)2] = n² + n + 1. For n=10, 100 + 10 + 1 = 111.
Question 2 of 6hard
If x, y, z are positive real numbers, the minimum value of (x+y)(y+z)(z+x) is:
8xyz
4xyz
6xyz
xyz
Explanation: Using AM ≥ GM for each pair: x+y ≥ 2√(xy), y+z ≥ 2√(yz), z+x ≥ 2√(zx). Multiplying gives (x+y)(y+z)(z+x) ≥ 8√(x²y²z²) = 8xyz.
Question 3 of 6hard
If the roots of the equation x³ − 12x² + 39x − 28 = 0 are in AP, then the common difference is:
1
2
3
4
Explanation: Let roots be a−d, a, a+d. Sum = 3a = 12 → a=4. Product = (4−d)(4)(4+d) = 28 → 16 − d² = 7 → d² = 9 → d = 3.
Question 4 of 6hard
The value of 2² + 4² + 6² + ... + 20² is:
1540
1440
1640
1500
Explanation: The series is (2×1)² + (2×2)² + ... + (2×10)² = 4(1² + 2² + ... + 10²) = 4 × 385 = 1540.
Question 5 of 6hard
If the sum of an infinite GP is 12 and r = 1/4, then the first term is:
9
8
7
6
Explanation: a/(1−1/4)=12 ⇒ a=9.
Question 6 of 6hard
The interior angles of a polygon are in AP. The smallest angle is 120° and the common difference is 5°. The number of sides of the polygon is:
9
12
16
Both 9 and 16
Explanation: Sum of angles = (n/2)[240 + (n − 1)5] = 180(n − 2). Solving gives n = 9 or 16. If n=16, the largest angle is 120 + 15(5) = 195° (invalid for a convex polygon). So, n = 9.
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