Question

if the number 2n-1,3n+2 and 6n-1 are AP. find n and hence the number

Answer 1

t1=2n-1
t2=3n+2
t3=6n-1
in ap t2-t1 = t3 -t2
(3n+2) -(2n-1)= (6n-1)-(3n+2)
n+3= 3n-3
3+3= 3n-n
2n=6
n=3
numbers are 2n-1 = 2(3)-1 =5
3n+2 = 3(3)+2 =11
6n-1 = 6(3)-1 = 17
n=3 and numbers are 5,11,17 which are in ap

Answer 2

HEY BUDDY...!!!

HERE IS UR ANSWER.

________________________

▶️If the number 2 n - 1 , 3 n + 2 and 6 n - 1 are
AP.

⏺️ Then we know

=> ( 3 n + 2 ) - ( 2 n - 1 ) = ( 6 n - 1 ) - ( 3 n + 2 )

▶️ As , A2 - A1 = A3 - A2

=> 3 n + 2 - 2 n + 1 = 6 n - 1 - 3 n - 2

=> n + 3 = 3 n - 3

=> 3 n - n = 3 + 3

=> 2 n = 6

=> n = 6 / 2

=> n = 3

⏺️ So the numbers are

=> 2 n - 1 = 2 × 3 - 1 = 6 - 1 = [ 5 ]

=> 3 n + 2 = 3 × 3 + 2 = 9 + 2 = [ 11 ]

=> 6 n - 1 = 6 × 3 - 1 = 18 - 1 = [ 17 ]

⏺️ So the A.P. is 5 , 11 , 17

HOPE HELPED..

GOOD NIGHT..

:-)