Progression Quiz  Aptitude Important MCQ
Quiz solution given below the Post
 (A) 297
 (B) 296
 (C) 287
 (D) 295
 (A) n² :n+2
 (B) n:n+3
 (C) n:2^{(n+2) }
 (D) n:2^{(n+3)}
 (A) 1922
 (B) 1891
 (C) 1829
 (D) 1860
 (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
 (A) 5643
 (B) 5640
 (C) 2928
 (D) 5673
 (A) 20, 20/7, 180/14, ....
 (B) 40, 80/3, 160/9, ...
 (C) 8, 16/3, 32/9, ...
 (D) None of these
 (A) 16,8,4
 (B) 2,8,4
 (C) 4, 8, 20
 (D) 10, 16, 21
 (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
 (A) 23/50
 (B) 23
 (C) 50
 (D) 50/23
 (A) 8, 12, 14
 (B) 10, 12, 14
 (C) 12, 16, 16
 (D) 20, 22, 26
Download Progression MCQ Aptitude
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
 (A) 34
 (B) 38
 (C) 35
 (D) 36
 (A) 5
 (B) 5
 (C) 4
 (D) 4
 (A) 883
 (B) 887
 (C) 885
 (D) 889
 (A) 3 if n is odd
 (B) 0 if n is even
 (C) Both and
 (D) None of these
 (A) 99:89
 (B) 99:83
 (C) 99:80
 (D) 99:86
 (A) Rs.10.5
 (B) Rs.l4
 (C) Rs.8
 (D) Cannot be determined
 (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 + (621)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 + (201)d)= 210, 2a + 19d21Sum of 10th
and 11th term = a+9d + a+10d = 2a + 19d = 21.

5

Exp. Sum of AP =
(n/2)(2a + (n1)d).a = 3, d = 6  3 = 3.a + (n1)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/(1r) =
120.Sum to infinity of squares of the terms of GP = a^2 / (1  r^2) =
2880.a/(1r) 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 ad, a , a+d.ad + a + a+d = 36, a = 12.(ad)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 + (n1)d = 55.Sum of AP = (n/2)(a + a + (n1)d) = 720.(n/2)(5 + 55)
= 720.n = 24.a + (n1)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(r1) = 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)/(r1)=
7(1281)/(21) = 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 + (21)B =
A + B = 10.5.

19

Exp. a = 16.16³
= 7 x (16³/(1r³)  16³).r = 0.5.Sum to infinity =
a/(1r) = 16/(10.5) = 32.

20

