# How could you complete this sequence: 49, 62, 70, 77, 91?

Updated on : December 7, 2021 by Aidan Butler

## How could you complete this sequence: 49, 62, 70, 77, 91?

This kind of question is quite misleading.

We have to check the difference between series numbers and observe some pattern, through observations we can find the answer very easy.

The series is

49, 62,70,77,91, x

we have found the next term in the series = x Difference between 2 consecutive numbers

62-49 = 13 = (4 + 9)

70-62 = 8 = (6 + 2)

77-70 = 7 = (7 + 0)

91-77 = 14 = (7 + 7)

x-91 = 10 = (9 + 1) *****

We can also see in the series that the sum of the digit of a number is equal to the difference between the number.

By eq. ***** we find x = 101

This is the first question about quora that I am answering.

Solution: The sum of the digit of once and the digit of the tens is added to the number to obtain the next digit.

49 # 4 + 9 = 13, the next digit is 49 + 13 = 62

62 # 6 + 2 = 8, the next digit is 62 + 8 = 70

70 # 7 + 0 = 7, the next digit is 70 + 7 = 77

77 # 7 + 7 = 14, the next digit is 77 + 14 = 91

91 # 9 + 1 = 10, the next digit is 91 + 10 = 101

Thanks!

Have fun!

Rajendra

rajendra.choudhary@aecom.com

49,62,70,77,91,101,103,107,115 and so on.

The difference between two numbers is the sum of the digits of the previous number.

The answer is math \ pi / math.

The numbers you posted are part of something called the CM (cow dung) function, and it's defined as:

f (1) = 49

f (2) = 62

f (3) = 70

f (4) = 77

f (5) = 91

f (6) = pi

f (n) = 0 for n> 6

There are two series in this series

49,70,91 ....... here the difference is 21 so the following numbers are 112,133,154 ....

62,77 ...... the difference is 15, so the following numbers are 92,107,122

So the final sequence is 49,62,70,77,90,92,112,107 .....

49 62 70 77 91? (92)

49

62 62-49 = 13

70– 62 = 8

77-70 = 7

91-77 = 14

49 70 91 (Difference 21)

62 77 92 (Difference 15)

The next number is 92

The next one will be 101, since the next number is equal to the sum of the digits added to the number. So the number after 91 = 9 + 1 + 91 = 101.

Right now there

I almost hate adding another answer "56" and ruining the total.

There I was, the laziest vendor of all time, serving Saint Alfonzo's pancake breakfast for my friend, Father O'Blivion. My pancakes were such a light brown fluffy color that they were the best in town.

I cooked a pancake. I got bored and didn't feel like doing more. But he had 56 people to feed. To do?

I cut the pancake in half. Now I had two pancakes. I remembered a story Father O'Blivion told me about Jesus feeding two fish to a group of people. It gave me an idea of what to do with my two pancakes.

The first cut

Keep readingRight now there

I almost hate adding another answer "56" and ruining the total.

There I was, the laziest vendor of all time, serving Saint Alfonzo's pancake breakfast for my friend, Father O'Blivion. My pancakes were such a light brown fluffy color that they were the best in town.

I cooked a pancake. I got bored and didn't feel like doing more. But he had 56 people to feed. To do?

I cut the pancake in half. Now I had two pancakes. I remembered a story Father O'Blivion told me about Jesus feeding two fish to a group of people. It gave me an idea of what to do with my two pancakes.

The first cut cuts the pancake into 2 pieces. For n> 1, the nth cut crossed each previous cut (avoiding parallelism) and also avoided each previous cut intersection, thus increasing the number of pieces in n.

When I had zero cuts, I had a pancake. After making a cut, I ate two pancakes. After two cuts, I had four pancakes. After three cuts, I had seven pancakes. And so on…

math \ begin {array} {c | c} \ text {Number of cuts} & \ text {Pancake pieces} \\ \ hline 0 & 1 \\ 1 & 2 \\ 2 & 4 \\ 3 & 7 \\ 4 and 11 \\ 5 and 16 \ \ 6 and 22 \\ 7 and 29 \\ 8 and 37 \\ 9 and 46 \\ 10 and 56 \\ \ end {array} / math

And there it was, after 10 cuts there were enough pieces of pancake for all 56 people to have one.

Above: I am not an anarchist! I am a lazy caterer who cuts the pancake three times to make seven pieces.

I decided to call this sequence of numbers the "lazy catering sequence", in honor of myself. Oh, Saint Alphonsus would be proud of me!

Since you got over that answer, I put this here for your listening pleasure after breakfast:

Original question: "What is the next number in the series, 1, 2, 4, 7, 11, 16, 22, 29, 37, 46, __?"

To find out the missing number, you need to find out what kind of pattern this is.

Since the series you are describing is one of the simplest (debatable), that is, an arithmetic series, I would not try to give examples for all kinds of series in mathematics.

Some common types to consider:

- Arithmetic
- Geometric
- Oscillatory
- Others more complex like logarithmic and exponential….

In the case of arithmetic series if we subtract the 2nd term (81) from the 1st term (83) we realize that the difference is 2. This is the same for the 3rd and 2nd, 4th and 3rd and so on ... There will be A DIFFERENT COMMON

Keep readingTo find out the missing number, you need to find out what kind of pattern this is.

Since the series you are describing is one of the simplest (debatable), that is, an arithmetic series, I would not try to give examples for all kinds of series in mathematics.

Some common types to consider:

- Arithmetic
- Geometric
- Oscillatory
- Others more complex like logarithmic and exponential….

In the case of arithmetic series if we subtract the 2nd term (81) from the 1st term (83) we realize that the difference is 2. This is the same for the 3rd and 2nd, 4th and 3rd and so on ... There will be a DIFFERENCE COMMON.

To find out what comes after 77 (fourth term) we need to subtract 2 to get 75. We can check this to see that 73 (the next term) is 2 plus less than 75.

In the case of geometric series instead of subtraction, we use division to find a COMMON RELATIONSHIP. Try to see if you can figure out how the missing number in a geometric series is supposed to be found! Greetings: D

What is the next number in the following sequence, 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, _?

This set of numbers is simply

number, number added to itself, next number, next number added to itself, ...

The pattern is easier to recognize with a small space between the pairs, like this:

1,2,2,4,3,6,4,8,5,10,

so starting with number 1 ...

1,

then adding 1 to itself (1 + 1) = 2,

1, 2

next number (number after 1) = 2

1,2, 2

then adding 2 to itself (2 + 2) = 4,

1.2, 2.4

next number (number after 2) = 3

1,2,2,4,3

then adding 3 to itself (3 + 3) = 6

1.2, 2.4, 3.6

next number (number after 3) = 4

1.2, 2.4, 3.6, 4

then adding 4 ai

Keep readingWhat is the next number in the following sequence, 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, _?

This set of numbers is simply

number, number added to itself, next number, next number added to itself, ...

The pattern is easier to recognize with a small space between the pairs, like this:

1,2,2,4,3,6,4,8,5,10,

so starting with number 1 ...

1,

then adding 1 to itself (1 + 1) = 2,

1, 2

next number (number after 1) = 2

1,2, 2

then adding 2 to itself (2 + 2) = 4,

1.2, 2.4

next number (number after 2) = 3

1,2,2,4,3

then adding 3 to itself (3 + 3) = 6

1.2, 2.4, 3.6

next number (number after 3) = 4

1.2, 2.4, 3.6, 4

then adding 4 to itself (4 + 4) = 8

1.2, 2.4, 3.6, 4.8 ...

You get the picture.

So the answer to the question "What is the next number in the following sequence, 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, _?"

The next number would be 6 followed by 12, then 7, 14 then 8, 16 ...

It is really difficult to determine the last four numbers of that sequence. The pattern is:

math \ frac {1} {2} x ^ 2 - \ frac {21} {2} x + 110 / math, where math x = 1..6 / math.

If we extend math x / math to math x = 1..10 / math, the last four numbers are:

math 61 / math

math 58 / math

math 56 / math

math 55 / math

Starting with math x = 11 / math, the sequence is reversed and begins to enlarge. If we extend math x / math to math x = 1..20 / math, the last four numbers are the same as the first four numbers in their sequence, but in reverse order since the sequence is reversing:

math 76 / math

math 83 / math

math 91 / math

math 100 / math

If we extend math x / math to math x = 1..30 / math, the last four numbers are:

mathematics 191 / mathematics

mathematics 208 / mathematics

mathematics 226 / mathematics

mathematics 245 / mathematics

If we extend math x / math to math x = 1..100 / math, the last four numbers are:

mathematics 3796 / mathematics

mathematics 3, / mathematics

It is really difficult to determine the last four numbers of that sequence. The pattern is:

math \ frac {1} {2} x ^ 2 - \ frac {21} {2} x + 110 / math, where math x = 1..6 / math.

If we extend math x / math to math x = 1..10 / math, the last four numbers are:

math 61 / math

math 58 / math

math 56 / math

math 55 / math

Starting with math x = 11 / math, the sequence is reversed and begins to enlarge. If we extend math x / math to math x = 1..20 / math, the last four numbers are the same as the first four numbers in their sequence, but in reverse order since the sequence is reversing:

math 76 / math

math 83 / math

math 91 / math

math 100 / math

If we extend math x / math to math x = 1..30 / math, the last four numbers are:

mathematics 191 / mathematics

mathematics 208 / mathematics

mathematics 226 / mathematics

mathematics 245 / mathematics

If we extend math x / math to math x = 1..100 / math, the last four numbers are:

math 3796 / math

math 3883 / math

math 3971 / math

math 4060 / math

If we expand math x / math to math x = 1..1000 / math (thousand), the last four numbers are:

mathematics 486,646 / mathematics

mathematics 487,633 / mathematics

mathematics 488,621 / mathematics

mathematics 489,610 / mathematics

If we extend math x / math to math x = 1..10000 / math (ten thousand), the last four numbers are:

math 49,865,146 / math

math 49,875,133 / math

math 49,885,121 / math

math 49,895,110 / math

If we extend math x / math to math x = 1..1000000 / math (one million), the last four numbers are:

math 499,986,500,146 / math

math 499,987,500,133 / math

math 499,988,500,121 / math

math 499,989,500,110 / math

If we expand math x / math to math x = 1..1000000000 / math (billion), the last four numbers are:

math 499,999,986,500,000,146 / math

math 499,999,987,500,000,133 / math

math 499,999,988,500,000,121 / math

math 499,999,989,500,000,110 / math

I imagine we could keep extending math x / math to infinity. Perhaps you should have asked, "What are the next four numbers in this sequence: 100, 91, 83, 76, 70, 65, __, __, __, __?" So we could answer with something simple like:

math 61 / math

math 58 / math

math 56 / math

math 55 / math

Original question: "What are the last four numbers in this sequence: 100, 91, 83, 76, 70, 65, __, __, __, __?"

A big thanks to Robert Tamburo for asking 901 questions like this (so far).

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