Q64.If a, b, c are in A.P. and a2, b2, c2 are in G.P. such that a < b < c and a + b + c = 34 , then the value of a is JEE Main 2018 (15 Apr Shift 2 Online) JEE Main Previous Year Paper (1) 1 4 − 3√21 (2) 14 − 4√21 (3) 1 (4) 1 1 − 4 √2 4 − 2√21
What This Question Tests
This question tests the ability to set up and solve a system of equations based on the definitions of arithmetic and geometric progressions, ultimately leading to a quadratic equation.
Concepts Tested
Formulas Used
2b = a + c
b² = ac
📚 NCERT Sections This Tests
3.21 — The Following Data Were Obtained During The First Order Thermal
Chemistry Class 11 · Chapter 3
3.21 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2 Cl 2 g SO 2 g Cl 2 g Experiment Time/s–1 Total pressure/atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm.
3.10 — In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 · Chapter 3
3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
3.20 — For The Decomposition Of Azoisopropane To Hexane And Nitrogen At 543
Chemistry Class 11 · Chapter 3
3.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant.
📋 Question Details
- Chapter
- Sequences & Series
- Topic
- Arithmetic and Geometric Progression
- Year
- 2018
- Shift
- 15 Apr Shift 2 Online
- Q Number
- Q64
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 9: Sequences & 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