Question

7,19,37,61,91....... is this a sequence if yes mention the type of the sequence.

Answer 1

Doesn't look like a sequence to me...

Answer 2

my i call it hexagonal centred numbers sequence

Answer 3

i just look that up.
yes, thats correct (even though the '1' in the beginning is missing.)

Answer 4

3n (n-1)+1

Answer 5

+12, +18, +24, +30 are the differences... which yes looks hexagonal: http://www.drking.org.uk/hexagons/misc/numbers.html

Answer 6

yes, 2nd differences are constant +6
so the n'th term had to be quadratic.

Answer 7

without 1 may it be hexagonal centred numbers ?

Answer 8

yes, yes,
just don't forget to mention that 'n' cannot take the value of 1
an = 3n(n-1)+1 for n=2,3,4....

Answer 9

find 4th term of the give sequence an show me it is correct

Answer 10

for 2st term, put n=2, so for 4th term put n=5
but thats not the usual way, so we need to edit the formula

an = 3(n+1)n +1
now put n=4 for 4th term

Answer 11

Replace n with n+1

an = 3(n+1)(n)+1 for n=1,2,3...

Answer 12

for 1st term**

Answer 13

so what is the 4th term according to your formula

Answer 14

Replacing n with n+1 skips the first term, and yields 7,19,37,61,91

Answer 15

4th term ----> n=4
a4 = 3*4*5+1 = 61

Answer 16

replacing n with n+1 is right or wrong

Answer 17

why that doubt ?
we initially had 3n(n-1)+1 but n started from 2.
to make it start from 1, we replaced 'n' by 'n+1' (yes, we can)
so we get an = 3n(n+1)+1 , n=1,2,3....

Answer 18

Why do you keep doubting us? Why not check our formulas for yourself to see if they're valid...?

if n=1, then n+1 = 2. Thus, replacing n with n+1 skips the first term (which would be 1).

Answer 19

^ since as hartnn said, "for 1st term, put n=2, so for 4th term put n=5"