Practice question on sequence , series and logarithm (aptitude)

Sequence , Series and Logarithm

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Question no. 1

What is the eighth term of the sequence 1, 4, 9, 16, 25.... ?

looks_one 8

looks_two 128

looks_3 64

looks_4 None of these

Option looks_3 64

Solution :
Given series consists of the perfect square terms , so given sequence is 1 , 4 ,9 , 16 ,25 , 36 , 49 , 64 .....
so the 8th term is 64

Question no. 2

The value of (1³+2³+3³ +....+15³) – (1+2+3+.... 15) is:

looks_one 14280

looks_two 14400

looks_3 12280

looks_4 None of these

Option looks_one 14280

Solution :
For cubic power terms

(1³+2³+3³ +....+15³) – (1+2+3+.... 15) = 14400 - 120 = 14280

Question no. 3

Iflog_a b= 1/2, log_bc= 1/3 and log_c a = K/5 , then the value of K

looks_one 25

looks_two 35

looks_3 20

looks_4 30

option looks_4 30

Solution :
log_ab = 1/2
log_bc = 1/3
log_ac = K/5
log_ab * log_bc = 1/6
log_ac = 1/6
log_ca = 6 = K/5
K = 6*5 = 30

Question no. 4

If a,b,c,d are in G.P., then (a³ + b³)^–1, (b³ + c³)^–1,(c³ + d³)^–1 are in

looks_one AP

looks_two GP

looks_3 HP

looks_4 None of these

Option looks_two GP

Solution :
b = ar , c = ar² , d = ar³

Question no. 5

If the sum of the series 54 + 51 + 48 + ........... is 513, then the number of terms are

looks_one 18

looks_two 20

looks_3 17

looks_4 None of these

Option looks_one 18

Solution :
Given series is in AP
a = 54 , d = -3

Question no. 6

If log2 = 0.3010 then the Arithmetic mean of log 40 and log 5 is

looks_one 2.301

looks_two 30.10

looks_3 3.0103

looks_4 1.1505

Option looks_4 1.1505

Solution :
log2 = 0.3010
AM = (log40 + log5)/2
= (log200)/2
= (log2 + log100)/2
= (.301 + 2 ) / 2 = 1.1505

Question no. 7

Which of the following are the two geometric means between 686 and 2 ?

looks_one 99 , 1

looks_two 98 , 14

looks_3 98 , 15

looks_4 None of these

Option looks_two 98 , 14

Solution :
2 , G₁ , G₂ , 686
a = 2
ar³ = 686
r³ = 686/2 = 343 ⇒ r = 7
G₁ = ar = 2 * 7 = 14
G₂ = ar² = 2 * 7² = 98

Question no. 8

If log2 = 0.30103, then find the number of digits in 2⁵⁶.

looks_one 13

looks_two 15

looks_3 17

looks_4 19

Option looks_3 17

Solution :
log2 = 0.30103
log₁₀56 = 56 * 0.30103 = 16.85
it means that this number has 16 zeros equivalent number
therefore it has 17 digits.

Question no. 9

Find three numbers in an arithmetic sequence such that the sum of the first and third is 12 and the product of the first and second is 24.

looks_one 4 ,6 , 8

looks_two 4 , 8 , 12

looks_3 4 , 5 , 6

looks_4 3 , 8 , 9

Option looks_one 4 ,6 , 8

Solution :
The terms are a - d , a , a + d
According to question
a - d + a + d = 12 ⇒ 2a = 12 ⇒ a = 6
(a-d)a = 24 ⇒ (6-d)6 = 24 ⇒ d = 2
first term = a-d = 6-2 = 4
second term = a = 6
third term = a + d = 6+2 = 8

Question no. 10

The pollution in a normal atmosphere is 0.01%. Due to leakage of a gas from a factory, the pollution is increased to 41%. If every day 50% of the pollution is neutralised, in how 51. many days the atmosphere will be normal ?

looks_one 10

looks_two 12

looks_3 20

looks_4 32

Option looks_two 12

Solution :
0.01 = 41(0.5)ⁿ
1/4100 = (1/2)ⁿ
n = 12

Question no. 11

If log₂[log₇(x²–x+37)]=1, then what could be the value x ?

looks_one 3

looks_two 5

looks_3 4

looks_4 None of these

Option looks_3 4

Solution :
log₂[log₇(x²–x+37)]=1
log₇(x²–x+37) = 2
x²–x+37 = 49
x²–x -12 = 0
x² - 4x + 3x -12 = 0
(x-4)(x+3) = 0
x = 4 , -3

Question no. 12

If log₁₀x – log₁₀√x = 2 log_x10 , then a possible value of x is given by

looks_one 10

looks_two 100

looks_3 1 /100

looks_4 None of these

Option looks_two 100

Solution :
log₁₀x – log₁₀√x = 2 log_x10

Question no. 13

If log₁₀ a +log₁₀ b = c , then the value of a is

looks_one bc

looks_two b/c

looks_3 10^c / b

looks_4 10b/c

Option looks_3 10^c / b

Solution :
log₁₀ a +log₁₀ b = c
log₁₀ ab = c
ab = 10^c
a = 10^c / b

Question no. 14

It is estimated that the population of a certain town will increase 10% each year for four years. What is the percentage increase in population after four years.

looks_one 50 %

looks_two 40 %

looks_3 42 %

looks_4 46 %

Option looks_4 46 %

Solution :
a= 1.1 (increase in first year)
r = 1.1 (Increase 10 % every year)
Increase after 4 year , ar³ = 1.1 * (1.1)³ = 1.46
Increase in percentage = (1.46 - 1)*100 % = 46 %

Question no. 15

The first term of a geometric sequence is 160 and the common ratio is 3/2. How many consecutive terms must be taken to give a sum of 2110?

looks_one 2

looks_two 4

looks_3 3

looks_4 5

Option looks_4 5

Solution :
a = 160
r = 3/2