closed
thats a very good start ! so, u got \(r = 3\)
1st term = 1 = 3^0 2nd term = 3 = 3^1 3rd term = 9 = 3^2 4th term = 27 = 3^3 .... 15th term = ?
nope
try again
try again
stare a bit at the pattern :)
1st term = 1 = 3^0 2nd term = 3 = 3^1 3rd term = 9 = 3^2 4th term = 27 = 3^3 .... 15th term = 3^14
\(\huge 3^{14}\) is your 15th term
3^14 = 4782969
yes
use below formula : \(n\)th term for arithmetic sequence : \(\large a_n = a + (n-1)d\) \(a\) = first term \(d\) = common difference
-7, -3, 1, 5, 9, \(a = -7\) \(d = 4\)
so, the 1000th term : \(\large a_{1000} = -7 + (1000-1)*4\)
simplify
remember below : ##arithmetic sequence : if first term = \(a\) common difference = \(d\) then, \(n\)th term is given by : \(\color{red}{a_n = a + (n-1)d}\) ##geometric sequencce : if first term = \(r\) common ratio = \(r\) then, \(n\)th term is given by : \(\color{red}{a_n = ar^{n-1}}\)
why did u delete replies lol :/
wat did u get after simplifying ?
u shouldnt get a really long number
\(\large a_{1000} = -7 + (1000-1)*4 \) \(\large ~~~~~~~~= -7 + (999)*4 \) \(\large ~~~~~~~~= -7 + 3996\) \(\large ~~~~~~~~= 3989\)
you got that ha ?
@iappreciateyourhelp
may i knw the really long answer u got ha ?b
okay fine
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