Question

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

Answer 1

Given: (2n−1),(3n+2) and (6n−1) are in AP.

So, difference between two consecutive terms is same.

So, (3n+2)−(2n−1)=(6n−1)−(3n+2)

(3n+2)+(3n+2)=6n−1+2n−1

6n+4=8n−2

8n−6n=4+2

or n=3

Numbers are:

2×3−1=5

3×3+2=11

6×3−1=17

Hence,(5,11,17) are required numbers.

Answer 2

Answer:

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