Progression - Aptitude Questions and Answers

 Arithmetic Progression (AP): A sequence where each term increases or decreases by a constant difference.

Geometric Progression (GP): A sequence where each term is multiplied by a constant ratio (r).

Harmonic Progression (HP): A sequence where reciprocals of terms form an AP.

Progression questions and answers are provided below for you to learn and practice.

Question 1: Find the nth term for the AP: 11, 17, 23, 29, ...

Solution: 

Here, a = 11, d = 17 - 11 = 23 - 17 = 29 - 23 = 6
We know that nth term of an AP is a + (n - 1) d
nth term for the given AP = 11 + (n - 1) 6
nth term for the given AP = 5 + 6 n
We can verify the answer by putting values of 'n'.
n = 1 -> First term = 5 + 6 = 11
n = 2 -> Second term = 5 + 12 = 17
n = 3 -> Third term = 5 + 18 = 23 and so on ... 

Question 2: Find the sum of the AP in the above question till the first 10 terms.

Solution : 

From the above question,
nth term for the given AP = 5 + 6 n
First term = 5 + 6 = 11
Tenth term = 5 + 60 = 65
Sum of 10 terms of the AP = 0.5 n (first term + last term) = 0.5 x 10 (11 + 65)
Sum of 10 terms of the AP = 5 x 76 = 380 

Question 3: For elements 4 and 6, verify that A ≥ G ≥ H.

Solution : 

A = Arithmetic Mean = (4 + 6) / 2 = 5G = Geometric Mean = \sqrt{{4}\times{6}}      = 4.8989H = Harmonic Mean = (2 x 4 x 6) / (4 + 6) = 48 / 10 = 4.8
Therefore, A ≥ G ≥ H 

Question 4: Find the sum of the series 32, 16, 8, 4, ... upto infinity.

Solution : 

First term, a = 32Common ratio, r = 16 / 32 = 8 / 16 = 4 / 8 = 1 / 2 = 0.5
We know that for an infinite GP, Sum of terms = a / (1 - r)
Sum of terms of the GP = 32 / (1 - 0.5) = 32 / 0.5 = 64 

Question 5: The sum of three numbers in a GP is 26 and their product is 216. ind the numbers.

Solution: 

Let the numbers be a/r, a, ar.
(a / r) + a + a r = 26
a (1 + r + r²) / r = 26
Also, it is given that product = 216
(a / r) x (a) x (a r) = 216
a³ = 216
a = 6
6 (1 + r + r²) / r = 26
(1 + r + r²) / r = 26 / 6 = 13 / 3
3 + 3 r + 3 r² = 13 r
3 r² - 10 r + 3 = 0
(r - 3) (r - (1 / 3) ) = 0
r = 3 or r = 1 / 3
Thus, the required numbers are 2, 6 and 18.

Question 6: Find the middle term of the A.P. 6, 13, 20, … , 216.

Solution:

In the given AP,

First term a₁ = 6

Last term a_n = 216

Common difference = 7

Now, to find the number of terms, n = (a_n - a₁)/d + 1
n = (216 - 6)/7 + 1
n = 210/7 + 1
n = 30 + 1
n = 31
So, middle term is (n + 1) / 2
= (31 + 1)/2
= 16
Now to calculate the middle term

a₁₆ = a1 + 15×d
a₁₆ = 6 + 15 × 7
a₁₆ = 6 + 105 = 111
So the middle term of the given AP is 111

Question 7: In an AP, if ? = −2, ? = 5 and ?? = 0, then find the value of ? ?

Solution:

Given values are:

d = -2

n = 5

a_n = 0

So, to calculate follow these steps:

a_n = a + (n-1)d

0 = a + (5-1)(-2)

0 = a - 8

a = 8

Question 8: Number of bacteria in a dish are 100, and they are increasing by double the previous value every hour. Find the number of bacteria in the dish after 6 hours. 

Solution:

Final amount=Initial amount × (growth factor)number of time periods

Here:

Initial amount = 100

Growth factor = 2 (since it's doubling)

Number of time periods (hours) = 6

So:

Number after 6 hours=100×2⁶
2⁶=64
100×64=6,400

Thus that after 6 hours, there are 6,400 bacteria.