Question

if the numbers n-1,n+3 and 3n -2are in AP, then the values of n is​

Answer 1

Answer

Step-by-step explanation:

So, difference between two consecutive terms is same

So,

=>(n+3)-(n-1)=(3n-2)-(n+3)

=>n+3-n+1=3n-2-n-2

=>4=2n-5

=>n=9/2

if wrong pls tell me

Answer 2

Answer:

Given : (n - 1), (n + 3) and (3n - 2) are in A.P.

So, difference between two consecutive terms is same.

So,

$(n + 3) - (n - 1) = (3n - 2) - (n + 3) \\ n + 3 - n + 1 = 3n - 2 - n - 3 \\ 4 = 2n - 5 \\ 4 + 5 = 2n \\ 9 = 2n \\ \boxed{ \bf \underline{ \underline{ \frac{9}{2} }} = n}$

Numbers are :

$n - 1 \rightarrow \frac{9}{2} - 1 = \frac{9 - 2}{2} = \frac{7}{2} \\ \\ n + 3 \rightarrow \frac{9}{2}+ 3 = \frac{9 + 6}{2} = \frac{15}{2} \\ \\ 3n - 2 \rightarrow 3( \frac{9}{2}) - 2 = \frac{27}{2} - 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{27 - 4}{2} = \frac{23}{2} \\ \\ \boxed{\bf Numbers \: \: are : \frac{7}{2}, \: \: \frac{15}{2} , \: \: \frac{23}{2} }$