2,2,10,47,150,?, 2970 find the missing number
Answers
Answer:
here you go !
Step-by-step explanation:
Thanks for Giving me an Opportunity to answer this wonderful Question !
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We have given a verbal series 2 , 2 , 10 , 47 , 150 , ? , 2970 .
We need to find the missing term of the series ?
Let's try to find the missing term of the series in how many possible ways possible to us ??
Method : 1
Given :
2, 2, 10 , 47 , 150 , ? , 2970
Solution :
Equation (1) :-
Equation (2) :-
Equation (3) :-
8 + 2
=> 10
Equation (4) :-
50 - 3
=> 47
Equation (5) :-
144 + 6
=> 150
Equation (6) :-
2401 + 49
=> 2450
Equation (7) :-
2916 + 54
=> 2970
Explanation :
Let's make group of terms for a Continuous series
Group 1 :- (2,2)
Group 2:- (10,47)
Group 3 :- (150 ,?)
Group :- (2970)
Note : The Power of 1 are Consecutive for all Groups
Start with Group 1
As you can see in Solution part we have made two equation ( Equation (1) and (2) respectively )
Here it's Solved in an Unique way
→ In the first equation the Power of 1 is use for the purpose of addition
Take 1 Common
1 + 1 = 2
(Here you can see one 1 extra . It is not the term but exponent of 1 used for the purpose of addition )
In the Second Equation also the power of 1 is use for the purpose of addition
→ 1^2 - 1^1 = 2
Take one Common
1 - 3
=> 2
Here we have added the power 1 to the power of 1 which is 2 (i.e. 1^(2 + 1) = 3)
Next is Group 2
{Refer to Equation (3) and (4) of the Solution part}
→ 1^3 + 2^3
We Know that ,
2^3 = 8
Here also we are going to use power with the purpose of addition
Subtract one from the power of 2 which is 3
2^(3-1)
2^2
Use Power for addition
2 + 8 = 10
→ 1^4 + 7^2
Have you Wondering why I used 7 instead of 3 ?
We need to add 5 with 2 for obtaining required or valid result
5 + 2
= 7
I used power of 2 as the power for 7
7^2
We have already go with Consecutive for 1 . So now , power of 1 is 4 (i.e. 1^4)
We know that ,
7^2 = 49
49 + 1 = 50 (Why?)
(7^2 + 1^4 = 49 + 1 = 50)
Use Power of exponent with the purpose of Subtraction
Subtract one to the power of 1 which is 4 (i.e. 1^4)
1^(4-1)
1^3
So , our Exponent is 3
50 - 3
=> 47
Next is for Group 3rd (i.e. 150 and the Unknown term )
→ 1^5 + 12^2
Here We used the Same pattern what we have used with 7
Adding of 5 with our previous term
7 + 5
= 12
Take 12^2
We Know that ,
12^2 = 144
144 + 1 = 145
Remain the Power of 1
Use Exponent of 1 with the purpose of addition
145 + 5
= 150
→ 1^6 + 49^2
Have you think why I used 49 ?
I added 37 with 12 ( previous term )
37 + 12
49 => Obtained Number
We Know that ,
49^2 = 2401
In this Equation and the next Equation we are not going with exponent . We Just have to add the number which we have obtained .
2401 + 49
=> 2450
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Going with our last Group (i.e. 2970)
{Refer to Equation 7}
1^7 + 54^2
In this we used 54 because 49 + 5 = 54
54 is our obtained result
We Know that ,
54^2 = 2916
Add the obtained result with 2916
2916 + 54
=> 2970
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Note for the Reader : I know well many of the reader confusing in equation part of my answer . But I want to say read once explanation part . I used exponent with purpose of either addition or Subtraction . Thanks!
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