Progression Quiz

1 >>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

Solution 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.

2 >>Q4. The sum to infinity of the series 1, 2, 4, 8, is _____. ?

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

Solution Exp. Sum to infinity exists only for a descending GP.

3 >>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

Solution 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.

4 >>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

Solution 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.

5 >>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

Solution 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.

6 >>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

Solution Exp. Only three such three digit numbers exist, 124, 139, 248.

7 >>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

Solution 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.

8 >>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

Solution 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.

9 >>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

Solution 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.

10 >>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

Solution 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.

11 >>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

Solution 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.

12 >>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 Exp. a = 16.16³ = 7 x (16³/(1-r³) - 16³).r = 0.5.Sum to infinity = a/(1-r) = 16/(1-0.5) = 32.

13 >>Q23. How many numbers between 450 and 950 are divisible by both 3 and 7? ?

• (A) 35
• (B) 20
• (C) 24
• (D) 30

Solution Exp. Numbers between 450 and 950 are divisible by both 3 and 7 = (950-450)/(3x7) = 24.

14 >>Q24. Sushant spent in the following manner. On the first day a month, he spent Rs. 100. On the second day of the month, he spent Rs. 200. On the third day of the month, he spent Rs. 400. If he continued to spend in this manner, on what day his total spending in that month exceeds Rs. 1.1 lakhs? ?

• (A) 12th
• (B) 8th
• (C) 10th
• (D) 11th

Solution Exp. Sum of GP = 1.1 lakhs.a(r^n - 1)/(r-1) = 1,10,000.100(2^n - 1)/(2-1) = 1,10,000.n > 10, hence on 11th day total expenditure exceeds 1.1 lakhs.

15 >>Q25. A square T2 is formed by joining the mid points of the sides of another square T1 of the side 16 cm. A third square is formed by joining the mid points of the sides of T2. What will be the sum of the areas of all the squares formed by repeating this process infinity, including area of T1 ? ?

• (A) 864 cm²
• (B) 512 cm²
• (C) 1024 cm²
• (D) None of these

Solution Exp. Area of T2 is half of T1, area of T3 is half of T2.The areas of the squares form a GP.Area of T1 = 16 x 16 = 256.The GP is, 256, 128, 64, 32,...., with common ratio as 0.5.Sum of areas to infinity = 256/(1-0.5) = 512.

16 >>Q26. The sum of first four terms of a G.P. is 120. The sum of its first two terms is 12. What will be the first term, if all its terms are (+ ve) ? ?

• (A) 5
• (B) 4
• (C) 3
• (D) 2

Solution Exp. a(r^4 - 1)/(r-1) = 120, a/(r-1) = 120/(r^4 - 1).a(r² - 1)/(r-1) = 12, a/(r-1) = 12/(r² - 1).120/(r^4 - 1) = 12/(r² - 1).r = 3.a = 3.