Derive s n th =u+a(n-1/2)
Answers
Answer:
Answer:s
Answer:s nth
Answer:s nth
Answer:s nth =μ+
Answer:s nth =μ+ 2
Answer:s nth =μ+ 2a
Answer:s nth =μ+ 2a
Answer:s nth =μ+ 2a (2n−1)
Answer:s nth =μ+ 2a (2n−1)The real formula is ;
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 )
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as times
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ]
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T]
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]a
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa (1)[2n−1] = LT
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa (1)[2n−1] = LT −2
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa (1)[2n−1] = LT −2 [ T ] [ T ]
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa (1)[2n−1] = LT −2 [ T ] [ T ] = L
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa (1)[2n−1] = LT −2 [ T ] [ T ] = L [ L ] = [ L ] + [ L ]
Answer:s nth =μ+ 2a (2n−1)The real formula is ; μ(1)+ 2a (1) (2n - 1 ) where 1 represent 1 sec as timesLHS = [ L ][ L ] = μ(1)=[LT −1 ][T] = [ L ]aa (1)[2n−1] = LT −2 [ T ] [ T ] = L [ L ] = [ L ] + [ L ] hence the formula is correct