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Progression Quiz | Aptitude Important MCQ

Quiz solution given below the Post

1 >>Q1. How many numbers are there in between 55 and 4505 which are multiples of 5 and divisible by 3? ?

  • (A) 297
  • (B) 296
  • (C) 287
  • (D) 295
2 >>Q2. The ratio of the n th terms of two arithmetic progressions is (2n - 1): (4n + 2). What is ratio of their sums to n terms ? ?

  • (A) n² :n+2
  • (B) n:n+3
  • (C) n:2(n+2)
  • (D) n:2(n+3)
3 >>Q3. The sum of the 3rd term and the 60th term of an P. is 60. If it has 62 terms. What will be the sum of all its terms? ?

  • (A) 1922
  • (B) 1891
  • (C) 1829
  • (D) 1860
4 >>Q4. The sum to infinity of the series 1, 2, 4, 8, is _____. ?

  • (A) Does not exist
  • (B) 1
  • (C) 0
  • (D) -1









5 >>Q5. The sum of first twenty terms of an arithmetic progression is 210. What will be the sum of its 10th and 11th term? ?

  • (A) 56
  • (B) 28
  • (C) 21
  • (D) 42
6 >>Q6. In a motorcycling race, initially 3 members are disqualified then 6 and then 9 and so on. At a certain particular stage, 183 members are disqualified. What will be the total number of motor cyclist disqualified in the race? ?

  • (A) 5643
  • (B) 5640
  • (C) 2928
  • (D) 5673
7 >>Q7. The sum of infinity of the terms of a G.P. is 120. The sum to infinity of the squares of the terms of the same G.P. is 2880. What will be the series? ?

  • (A) 20, 20/7, 180/14, ....
  • (B) 40, 80/3, 160/9, ...
  • (C) 8, 16/3, 32/9, ...
  • (D) None of these
8 >>Q8. What will be the three numbers in G.P. having a sum 28 and a product as 512? ?

  • (A) 16,8,4
  • (B) 2,8,4
  • (C) 4, 8, 20
  • (D) 10, 16, 21
9 >>Q9. How many three digit numbers have their digits distinct and in G.P. with an integral common ratio? ?

  • (A) 4
  • (B) 6
  • (C) 12
  • (D) 3





10 >>Q10. The sum of three numbers in an P. is 36. The sum of their squares is 440. What will be the largest among the three numbers? ?

  • (A) 18
  • (B) 16
  • (C) 14
  • (D) 12
11 >>Q11. The first term of an arithmetic progression is 5 and its last term is 55. If the sum of the terms of the P. is 720, the common difference is ________. ?

  • (A) 23/50
  • (B) 23
  • (C) 50
  • (D) 50/23
12 >>Q12. The sum of three numbers of P. is 36. If the sum of their squares is 440, the numbers are _______. ?

  • (A) 8, 12, 14
  • (B) 10, 12, 14
  • (C) 12, 16, 16
  • (D) 20, 22, 26


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13 >>Q13. A total of 255 coins is placed in n boxes, such that a person can take any number of coins from 1 to 255 by selecting a suitable combination of one or more than 1 boxes out of n boxes, the minimum value of n is ________________. ?

  • (A) 5
  • (B) 6
  • (C) 7
  • (D) 8
14 >>Q14. The 5th and 10th terms of an P. are 23 and 48 respectively. What will be its 8th term? ?

  • (A) 34
  • (B) 38
  • (C) 35
  • (D) 36
15 >>Q15. If there are 15 arithmetic means between 40 and 120, then the common difference of P. is ___________ ?

  • (A) 5
  • (B) -5
  • (C) -4
  • (D) 4
16 >>Q16. In a G.P. of positive terms, the difference of 9th and 8th terms is 896, and the difference of 2nd and 1st is 7. What will be the sum of first 7 terms of the series? ?

  • (A) 883
  • (B) 887
  • (C) 885
  • (D) 889
17 >>Q17. The sum to n terms of the G.P, -3, 3, -3, 3, -3, 3 n times is ___________ . ?

  • (A) -3 if n is odd
  • (B) 0 if n is even
  • (C) Both and
  • (D) None of these
18 >>Q18. The first term of an P. consisting of 30 terms is 10 and the common difference is 5. What will be the ratio of the sum of 30 terms of the progression and the sum of the last 20 terms of the progression? ?

  • (A) 99:89
  • (B) 99:83
  • (C) 99:80
  • (D) 99:86
19 >>Q19. Ankit and Ashok start saving money in a certain month of a year. Every month from then on, Ankit increased his savings by an amount equal to the savings of Ashok in the first month and Ashok increased his savings by an equal amount to the savings of Ankit in the first month. In 20 months the total of the amounts saved by both is equal to 2100, what will be the amount saved by Ankit in the second month? ?

  • (A) Rs.10.5
  • (B) Rs.l4
  • (C) Rs.8
  • (D) Cannot be determined
20 >>Q20. The first term of an infinite G.P. is 16. A new series is formed by cubing each term of the series. Now, each term of the new series is 7 times the sum of all the terms following it. What will be the sum to infintiy of the old progression? ?

  • (A) 32
  • (B) 18
  • (C) 21
  • (D) 24

Solution of Progression Quiz

1

2
Exp. Sum of 3rd term and 60th term= a+2d + a+59d = 60,2a + 61d = 60.Sum of 62 terms= (62/2)(2a + (62-1)d)= 31(2a+61d)= 31(60) = 1860.
3
Exp. Sum to infinity exists only for a descending GP.
4
Exp. Sum of first twenty terms of an P.= (20/2)(2a + (20-1)d)= 210, 2a + 19d21Sum of 10th and 11th term = a+9d + a+10d = 2a + 19d = 21.
5
Exp. Sum of AP = (n/2)(2a + (n-1)d).a = 3, d = 6 - 3 = 3.a + (n-1)d = 183, n = 61.Sum = (61/2)(2x3 + (60)3) = (61/2)(186) = 5673.
6
Exp. Let 'a' be the first term and 'r' be the common ratio.Sum to infinity of GP = a/(1-r) = 120.Sum to infinity of squares of the terms of GP = a^2 / (1 - r^2) = 2880.a/(1-r) x a/(1+r) = 2880.120 x a/(1+r) = 2880.a/(1+r) = 24.a = 40, r = 2/3.
7

8
Exp. Only three such three digit numbers exist, 124, 139, 248.
9
Exp. Let the three numbers be a-d, a , a+d.a-d + a + a+d = 36, a = 12.(a-d)² + a² + (a+d)² = 440.3a² + 2d² = 440.3(12²) + 2d² = 440, 2d² = 440 - 3(144).2d² = 8, d = 2.Largest number = a+d = 12+2 = 14.
10
Exp. Last term of AP = a + (n-1)d = 55.Sum of AP = (n/2)(a + a + (n-1)d) = 720.(n/2)(5 + 55) = 720.n = 24.a + (n-1)d = 55, 5 + 23d = 55, d = 50/23.
11

12

13
Exp. a + 4d = 23.a + 9d = 48, a + 4d + 5d = 48, 23 + 5d = 48, d = 5.a + 4x5 = 23, a = 3.8th term = a + 7d = 3 + 7x5 = 38.
14

15
Exp. Difference of 9th and 8th terms of a GP= a x r&sup8; - a x r&sup7;= a x r&sup7; x (r - 1)896Similarly, difference between 2nd and 1st term= a(r-1) = 7.From the two equations, r&sup7;= 896/7 = 128, r = 2.Hence, we have a = 7.Sum of first 7 terms of GP= a(r&sup7; - 1)/(r-1)= 7(128-1)/(2-1) = 896.
16

17
Exp. Sum of 30 terms = (30/2)(2x10 + 29x5) = 2475.Sum of first 10 terms = (10/2)(2x10 + 9x5) = 325.Sum of last 20 terms = 2475 - 325 = 2150Ratio = 2475/2150 = 99/86.
18
Exp. Let amount saved by Ankit in first month =Let amount saved by Ashok in first month =Sum saved by Ankit after 20 months= (20/2)(2A + 19B).Sum saved by Ashok after 20 months= (20/2)(B + 19A).Total sum= 10(A + 19B) + 10(B + 19A)= 200A + 200B = 2100.A + B = 10.5.Total amount saved by Ankit in second months= A + (2-1)B = A + B = 10.5.
19
Exp. a = 16.16³ = 7 x (16³/(1-r³) - 16³).r = 0.5.Sum to infinity = a/(1-r) = 16/(1-0.5) = 32.
20




 
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