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Mathematics 18 Online
danielfootball123:

HEEELLLPPP!!!!!!!! Find the sequence and fill in the blanks (2, 3), (4, 6), (6, 9), (8, 12) .... The 8th group is (__, __)

flowerpower52:

16;24

danielfootball123:

ARE YOU ONE HUNDRED % SURE

flowerpower52:

Yes

flowerpower52:

It’s doubling

danielfootball123:

THANKS GTG

danielfootball123:

BYE

danielfootball123:

THANKS FOR SAVING MY LIFE

flowerpower52:

Welks

flowerpower52:

@danielfootball123 wrote:
THANKS FOR SAVING MY LIFE
Did it work

Tbone:

It's not doubling. Well the y isn't. It's adding 3 each time. So it would technically be (16,15)

surjithayer:

2,4,6,8,... first term a=2 c.d.=4-2=2 \[t _{n}=a+(n-1)d\] put n=8 \[t _{8}=2+(8-1)2=2+14=16\] similarly second=24 so 8th term is (16,24)

Tbone:

@surjithayer wrote:
2,4,6,8,... first term a=2 c.d.=4-2=2 \[t _{n}=a+(n-1)d\] put n=8 \[t _{8}=2+(8-1)2=2+14=16\] similarly second=24 so 8th term is (16,24)
How did you get the Y tho?

Tbone:

Because for me it is very clearly not being doubled like the X is

Tbone:

It's being added by 3

Tbone:

But I am only in algebra 2 so idk

Aiden3j:

its 10 and 15

Aiden3j:

its going up by adding 2 each time on x and adding 3 each on y

jhonyy9:

@aiden3j wrote:
its going up by adding 2 each time on x and adding 3 each on y
i think it hence too ,this is the right way

jhonyy9:

@narad ?

Narad:

The nth term is \[u_{n}= n (2,3)= \left( 2n, 3n \right)\]

jhonyy9:

(2, 3), (4, 6), (6, 9), (8, 12) .... The 8th group is (__, __) sorry now i see that ask about 8th group ???

KyledaGreat:

To find the sequence, we can observe that the second coordinate of each pair is twice the first coordinate. Therefore, we can write the sequence as: (2, 3), (4, 6), (6, 9), (8, 12), ... To find the 8th group, we need to continue the pattern until we reach the 8th pair: (2, 3), (4, 6), (6, 9), (8, 12), (10, 15), (12, 18), (14, 21), (16, 24) Therefore, the 8th group is (14, 21).

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