infinite sequence

[yvonne12]

In the infinite sequence a,a1,a2,a3...an each term after the first is equal to twice the previous term. If a5-a2=12 what is the value of a1?

Please explain if this applies to arithmetic progression. How do I work this problem out?

[RAGS]

it is a geometric progression prob

a1 a2 a3 a4 a5 and each term is double the previous term so.......

a 2a 4a 8a 16a

no given is a5-a2=16a-2a=14a=12

ie a=6/7

[Cybermusings]

In the infinite sequence a,a1,a2,a3...an each term after the first is equal to twice the previous term. If a5-a2=12 what is the value of a1?

Please explain if this applies to arithmetic progression. How do I work this problem out?

This is not an arithmetic progression. In an arithmetic progression the difference between 2 consecutive terms of the sequence remains constant

This is not the case with this particular question.

This is a geometric progression. In a geometric progression the ratios between two consecutive terms of the sequence remains constant. Here a2/a1 = 2/1; a3/a2=2/1; so on and so forth

Now a5-a2 = 12

If a1 is x then a2 = 2x,a3=4x,a4=8x,a5=16x
So 16x - 2x = 12
14x=12
Hence x = 12/14 = 6/7
Thus the first term is 6/7

[yvonne12]

Thank you