Previous Year Topic Wise Questions With Solutions: Series | CSAT Preparation - UPSC PDF Download

Ans: (b)
The pattern is:
24 × 0.5 = 12
12 × 1 = 12 (X)
12 × 1.5 = 18
18 × 2 = 36
36 × 2.5 = 90
It appears that the multiplier is increasing by 0.5 each time. So, the missing term is 12.

Ans: (c)
Given sequence: A_BCD_BBCDABC_DABC_D The required sequence is:
A A BCD / A BBCD / ABC C D / ABC D D
Hence, the missing letters are – A, A, C, D

Ans: (b)
Given series: 3, 14, 39, 84, *, 258
The series follows the following pattern: 1³ + 1² + 1 = 3
2³ + 2² + 2 = 14
3³ + 3² + 3 = 39
4³ + 4² + 4 = 84
* = 5³ + 5² + 5 = 155
6³ + 6² + 6 = 258
Thus, missing number is 155.

Ans: (c)
tt0 is a 3-digit number ending with zero, such that pp + qq + rr = tt0
As pp, qq, and rr are 2-digit numbers, the value of tt0 can be either 110 or 220. Thus, we get:
If tt0 = 110
pp + qq + rr = 110
or (10p + p) + (10q + q) + (10r + r) = 110 
or 11p + 11q + 11r = 110
or p + q + r = 10 ……. (1)
If tt0 = 220
pp + qq + rr = 220
or (10p + p) + (10q + q) + (10r + r) = 220 
or 11p + 11q + 11r = 220
or p + q + r = 20 ……. (2)
Statement 1:
As p < q < r, the possible value of p in equation (1) can be 1 and 2.
As p < q < r, the possible value of p in equation (2) can be 1, 2, 3, 4 and 5.
Hence, the number of possible values of p is 5.
Thus, Statement 1 is correct.
Statement 2:
As p < q < r, the possible value of q in equation (1) can be 2, 3 and 4.
As p < q < r, the possible value of q in equation (2) can be 6, 7 and 8.
Hence, the number of possible values of q is 6.
Thus, Statement 2 is correct.
Hence, option (c) is correct.

Ans: (d)
The given series is: 20, 10, 10, 15, 30, 75, X?
The terms are decreasing in the initial half, and then they start increasing. The speed at which they increase at the latter half suggest that multiplication may be involved.
The pattern is as follows:
20 × 0.5 = 10
10 × 1 = 10
10 × 1.5 = 15
15 × 2 = 30
30 × 2.5 = 75
75 × 3 = 225

Ans: (c)
The given series is: AABABCABCDABCDE…
The pattern being followed here is: A, AB, ABC, ABCD, ABCDE, …
It’s kind of an arithmetic series wherein the first term has 1 letter, second term has 2 letters, and so on.
We have to estimate the length of the letter-series near the 100^th letter.
If we have an arithmetic series, 1, 2, 3 …, then:
Sum of first n terms = (n/2) (a + l), where ‘n’ is the number of terms in the series, ‘a’ the first term, and
‘l’ the last term.
So, sum of first 14 terms = (14/2) (1 + 14) = 7 × 15 = 105
So, the term having 14 letters will have the 100^th letter of the series.
It will start at the 92^nd letter and end at 105^th letter. Hence, the 100^th letter must be I.

Ans: (b)
The pattern has been represented below:
Hence, option (b) is the correct answer.

Ans: (a)
On separating the Odd and the Even Series, we observe the following pattern:

On observing the given two series, we observe that the first term in EVEN SERIES should be replaced by 0, therefore option (a) is the correct answer

Ans: (c)
The inherent pattern in the two sequences has been represented below. 

Ans: (d)
The given series is: _b_a_ba_b_abab_aab
The pattern being followed in the above series is: abb, then aab. Again abb, and then aab, and so on.
The blanks have been highlighted below:
a b b / a a b / a b b / a ab / ab b / aab

Ans: (c)
The given sequence is 2, 7, 22, 67, 202, X, 1822
The pattern being followed here is given below:
2 × 3 + 1 = 7
7 × 3 + 1 = 22
22 × 3 + 1 = 67
67 × 3 + 1 = 202
202 × 3 + 1 = 607
607 × 3 + 1 = 1822