Q64.Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square, whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is (1) 262 (2) 190 (3) 225 (4) 157
What This Question Tests
This question involves forming an equilateral triangle with balls, which translates to the sum of the first 'n' natural numbers. The problem then relates this sum to the number of balls in a square, requiring algebraic manipulation to find the original number of balls.
Concepts Tested
Formulas Used
Sn = n(n+1)/2
Area of square = s^2
📚 NCERT Sections This Tests
1.17 — A Point Charge +10 Mc Is A Distance 5 Cm Directly Above The Centre
Physics Class 11 · Chapter 1
1.17 A point charge +10 mC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.) FIGURE 1.31
2.2 — A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 · Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
6.11 — Dynamics Of Rotational
Physics Class 11 · Chapter 6
6.11 Dynamics of rotational the motion of extended bodies. motion about a fixed axis A large class of problems with extended bodies can be
📋 Question Details
- Chapter
- Sequences & Series
- Topic
- Sum of natural numbers
- Year
- 2019
- Shift
- 09 Apr Shift 2
- Q Number
- Q64
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 9: Sequences and Series
More from this Chapter
Q86.In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals (1) 1 2 (1 −√5) (2) 21 √5 (3) √5 (4) 12 (√5 −1)
Q88.The sum of the series 2! 1 −13! + 4!1 −… upto infinity is (1) e−2 (2) e−1 (3) e−1/2 (4) e1/2
Q71.Statement - 1: For every natural number n ≥2, 1 + 1 + … + 1 > √n. Statement −2 : For every √1 √2 √n natural number n ≥2, √n(n + 1) < n + 1. (1) Statement −1 is false, Statement −2 is true (2) Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1 (3) Statement −1 is true, Statement −2 is true; (4) Statement −1 is true, Statement −2 is false. Statement −2 is not a correct explanation for Statement −1.
Q76.The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is (1) −4 (2) −12 (3) 12 (4) 4