Question

2,2,10,47,150,?,2970 find the missing term

Answer 1

Hey Sir!

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) :- $1^1 + 1^1 = 2$

Equation (2) :-  $1^2- 1^1 = 2$

Equation (3) :- $1^3 + 2^3 = 8$

8 + 2

=> 10

Equation (4) :-  $1^4 + 7^2 = 1 + 49 = 50$

50 - 3

=> 47  

Equation (5) :-  $1^5 +12^2 = 144 + 1 = 144$

144 + 6

=> 150

Equation (6) :-  $1^6 + (49)^2 + 49$

2401 + 49

=> 2450

Equation (7) :-  $1^7 + (54)^2 + 54$

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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Warm Regards

@FuturePoet

Best User , Moderator

Answer 2

$\bold{\large{\underline{\underline{\sf{Given:}}}}}$

2, 2, 10 , 47 , 150 , ? , 2970

$\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}$

Eq. (1) :- 1¹ + 1¹ = 2

Eq. (1) :- 1¹ + 1¹ = 2Eq. (2) :- 1²- 1¹ = 2

Eq. (1) :- 1¹ + 1¹ = 2

Eq. (2) :- 1²- 1¹ = 2

Eq (3) :- 1³ + 2³ = 8 + 2 => 10

Eq. (4) :- 1⁴ + 7² = 1 + 49 = 50

50 - 3

=> 47

Eq.(5) :- 1^5 +12²= 144 + 1 = 144

144 + 6

=> 150

Eq. (6) :- 1^6 + (49)²+ 49\

Eq. (6) :- 1^6 + (49)²+ 49\2401 + 49

=> 2450

Equation (7) :- 1^7 + (54)^2 + 54

Equation (7) :- 1^7 + (54)^2 + 542916 + 54

=> 2970

So, your answer is 2450