Reliance Jio Airthmatic Progression Quiz 1

a = 4

 d = 7-4 =3 

 

a_{n} = a + (n-1) d = a +6d = 4 + 6\times 3 = 22

As we know about the nth term of AP is [a + (n-1)d]

So, 6th term = a + 5d

and 9th term = a + 8d

Given a + 5d = 56 ………(1)

a + 8d = 89……….(2)

Subtract (1) form (2)

3d = 36

d = 12

Solution

The series have multiple starting from 104 , 112 , 120 ………..200

It becomes an AP having common difference 8

Total number of natural numbers= \frac{[(\text{last term} – \text{first term})}{diff}] + 1

=\frac{[(200 – 104)]}{8} + 1

= 12 + 1 = 13

D = 9-5 =4

nth term = a + (n-1)d, 

41 = 5 + (n-1)3

 n = 13

Solution :

45 - 34 = 11

56 - 45 = 11

67 - 56 = 11

79 - 67 = 12

 

So it is not Arithmetic Progression

Solution 

A = 60

 d= 72 - 60 = 12 

 d =84 - 72 = 12

Ratio of a and d = \frac{60}12} = 5

Let First three terms are a- d , a , a +d 

 

Sum of first three terms = a- d+a +a+d = 3a 

 

According to question 

3a = 30

a = 10

n^{2} +4 =  5^{2} +4 = 29

Solution:

a = 24 

d = 35 - 24 = 11

13 th term = a +(n-1) d= 24 + (13-1)11 = 156

n^{2} +n + 2 =  10^{2} +10 + 2   = 112