This set of Aptitude Questions and Answers (MCQs) focuses on “Progression – Set 2”.
1. If a, b and c are in G.P., then 1/(1 + log10a), 1/(1 + log10b) and 1/(1 + log10c) will be in which of the following progression?
a) A.P.
b) G.P.
c) H.P.
d) Arithmetic Geometric Progression
View Answer
Explanation: Since a, b and c are in G.P., then b2 = ac
On taking logarithm to the base 10 on both sides, we get
2log10b = log10a + log10c
The above equation is in A.P., hence (1 + log10a), (1 + log10b) and (1 + log10c) will also be in A.P.
Therefore, 1/(1 + log10a), 1/(1 + log10b) and 1/(1 + log10c) are in H.P.
2. Find the sum of 1+\(\frac{3}{5}+\frac{5}{25}+\frac{7}{125}\)+⋯ up to n terms.
a) 15/8
b) 15/16
c) 25/16
d) 5/16
View Answer
Explanation: The given sequence is arithmetic geometric series, where a=1, d=2 and r=(1/5)
S∞ = \(\frac{a}{1-r} + \frac{dr}{(1-r)^2}\)
S∞ = \(\frac{1}{(1-1/5)} + \frac{2/5}{(1-\frac{1}{5})^2}\)
S∞ = 5/4 + 10/16 = 15/8
3. If a, 2a+2 and 3a+3 are in GP, then what is the fifth term of the GP?
a) -81/4
b) 81/4
c) -81/2
d) 81/2
View Answer
Explanation: Given a, 2a+2 and 3a+3 are in GP.
i.e., (2a+2)2 = a(3a+3)
a2 + 5a + 4 = 0
a = -1, -4
Now, a = -1 does not satisfy the given series.
Therefore, -4, -6 and -9 are in GP with r=3/2
Fifth term, T5 = ar4 = -4(3/2)4 = -81/4
4. Find the sum of the first n terns of the series \(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\)⋯.
a) 2n – n – 1
b) 1 – 2-n
c) n + 2-n – 1
d) 2n-1
View Answer
Explanation: \(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\)⋯ +n terms
= (1-\(\frac{1}{2}\))+(1-\(\frac{1}{4}\))+(1-\(\frac{1}{8}\)+⋯+(1-\(\frac{1}{2^n}\))
= n – (\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+⋯+\frac{1}{2^n}) \)
= n – \(\frac{1}{2}\Big(\frac{1-(\frac{1}{2})^n}{1-(\frac{1}{2})}\Big)\) = n + 2-n – 1
5. In a pond, flowers triple in every 6 minutes. If there are 3 flowers in the pond initially, then in what time the number of flowers become 177147?
a) 54 min
b) 60 min
c) 90 min
d) 4 min
View Answer
Explanation: Given that initially there are 3 flowers i.e., a=3 and Tn=177147
Since, the flowers triple itself, it is in GP with r=3
Tn = arn-1
177147 = 3*3n-1
n = 11
But time interval required is (n-1) i.e., 10
Therefore, time taken is 10*6 min = 60 minutes.
6. A ball is dropped from a height of 512 m. It bounces back rising to a height of 256 m. each time it further touches the floor it rises to the half the height it fell from before the previous bounce. Find the total distance travelled by the ball.
a) 1024
b) 2048
c) 1536
d) 768
View Answer
Explanation: The problem is of the kind infinitely long decreasing GP.
Given, a= 512 and r = ½
S∞ = \(\frac{a}{1-r}\) = 1024
The sum S∞ gives only the sum of falling height. It doesn’t take into account of bouncing back height.
Hence, for the sum of bouncing back height a=256, r=1/2 and S∞=512
Therefore, the total distance travelled is 1024+512=1536.
7. Given X = 365 and Y = (364+363+362+…+1). Which of the following is true?
a) Y is 364 larger than X
b) X and Y are equal
c) Y is larger than X by 1
d) X is larger than 2Y by 1
View Answer
Explanation: B is in GP with a=1=30, r=3 and n=65
Sn = Y = \(\frac{a(r^n-1)}{(r-1)} = \frac{1(3^{65}-1)}{3-1} \)
Y = (365 – 1)/2 = (X – 1)/2
2Y = X – 1
Therefore, X is larger than 2Y by 1
8. If log3, log(3x-1) and log(3x+5) are in AP, then what is the value of x?
a) 7/3
b) log35
c) log37
d) 5/3
View Answer
Explanation: Given log3, log(3x-1) and log(3x+5) are in AP
Hence, 2log(3x-1) = log3 + log(3x+5)
log(3x-1)2 = log[3*(3x+5)] 3x-1)2 = 3*(3x+5)
Let y = 3x
Hence, (y-1)2 = 3*(y+5)
y2 – 5y – 14 = 0
On factorizing, we get, y = 7 or y = -2
So, 3x = 7 or 3x = -2 (3x cannot be negative)
Hence, 3x = 7
Or x = log37
9. What is the sum of all 2-digit number that leaves a remainder of 1 when divided by 4?
a) 1210
b) 1220
c) 1120
d) 1212
View Answer
Explanation: The 2-digit number sequence which leaves a remainder of 1 when divided by 4 is 13, 17, 21, …., 97
Tn = a + (n-1)d
97 = 13 + (n-1)*4
n = 22
Sum of an AP, Sn = \(\Big(\frac{first \, term+last \, term}{2}\Big)\)*number of terms
S300 = \(\Big(\frac{13+97}{2}\Big)\)*22 = 1210
10. In an AP, the ratio of the 7th term to the 10th term is -1. If the 16th term is -15, what is the 3rd term?
a) 11
b) 13
c) -11
d) 13
View Answer
Explanation: Given T7/T10 = -1
\(\frac{a+6d}{a+9d}\) = -1
Hence, 2a = 15d …… (i)
Given, T16 = a + 15d = -15
a = 15 …… (from (i))
Therefore, d = -2 and T3 = a+2d = 11
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