Question

find expression for apparent nth term of geometric sequence 2,-4,6,-8,10,...

geometric sequence is first number times common ratio (ie. -2).
negative common ratio for alternating negative-postive sequence. But more is needed.

Answer 1

No way is this a geometric sequence.

Answer 2

I assumed that. arithmetic sequence?

Answer 3

not even arithmetic ... maybe harmonic??

Answer 4

how to find the expression for that sequence?

Answer 5

consider 2+4+6+8+10+12+...
first term=2...
c.d=4-2=2
\[tn=\left( -1 \right)^{n-1}\left\{ 2+\left( n-1 \right)2 \right\}=2\left( -1 \right)^{n-1}n\]

Answer 6

correction i have taken series it is a sequence

Answer 7

surjithayer, the sequence is 2,-4,6,-8,10

Answer 8

insert commas inplace of +

Answer 9

normally the expression would be first number * common ratio ^(n - 1)
alternating pos/neg indicates negative common ratio
but not in this case. I want to learn the steps if possible

Answer 10

\[\left( -1 \right)^02,\left( -1 \right)^14,\left( -1 \right)^26,\left( -1 \right)^38,\left( -1 \right)^410,....\]

Answer 11

interesting approach, but expression won't find nth term of sequence

Answer 12

\[2\left\{ \left( -1 \right)^{1-1}1,\left( -1 \right)^{2-1}2,\left( -1 \right)^{3-1}3,\left( -1 \right)^{4-1}4,\left( -1 \right)^{5-1}5,...,\left( -1 \right)^{n-1}n,... \right\}\]
\[tn=2\left( -1 \right)^{n-1}n\]

Answer 13

thank you!! but how can I derive that formula myself? The usual geometric formula is ar^(n-1)

Answer 14

no problem for 2,4,6,8,10,...
for powers of -1,terms are 0,1,2,3,4,....
a=0,d=1-0=1
tn=a+(n-1)d=0+(n-1)=n-1
so power is n-1

Answer 15

you got it.

Answer 16

thanks so much!!

Answer 17

yw