find n if n-2,4n-1and 5n+2
gaurav2013c:
Are they in AP??
this is in chapter ap
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Answered by
1
Since they are in AP
The common difference is constant
T3 - T2 = T2 - T1
=> 5n +2 - 4n+1 = 4n-1-n+2
=> n+3 = 3n +1
=> 3n-n = 3-1
=> 2n = 2
=> n= 1
The common difference is constant
T3 - T2 = T2 - T1
=> 5n +2 - 4n+1 = 4n-1-n+2
=> n+3 = 3n +1
=> 3n-n = 3-1
=> 2n = 2
=> n= 1
Answered by
0
average of a and c is equal to b.
Therefore, by the condition:
\frac{(n-2)+(5n+2)}{2}2(n−2)+(5n+2) =4n - 1.
⇒\frac{n-2+5n+2}{2}2n−2+5n+2 =4n -1
⇒\frac{6n}{2}26n =4n-1
⇒3n=4n-1
⇒4n-3n=1
⇒n=1
Therefore, by the condition:
\frac{(n-2)+(5n+2)}{2}2(n−2)+(5n+2) =4n - 1.
⇒\frac{n-2+5n+2}{2}2n−2+5n+2 =4n -1
⇒\frac{6n}{2}26n =4n-1
⇒3n=4n-1
⇒4n-3n=1
⇒n=1
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