# n AM's are inserted between 2 and 38. If third AM is 14, then n is equal to Option 1) 9 Option 2) 7 Option 3) 8 Option 4) 10

As we learnt in

Inserting n AMs between a and b -

$a_{1},A_{1},A_{2},A_{3},A_{4},- - - - A_{n},b$ are in AP

- wherein

a is the first term

b is the (n+2)th term of the AP.

$\\2,A_1,A_{2},A_{3},.......A_{n},38\\*\\*38=2+(n+1)d\\*\\*Also\, \, 14=2+3d= > d=4\\*\\*38=2+\left ( n+1 \right )4\\*\\*36=\left ( n+1 \right )4= > n=8$

Option 1)

9

Incorrect

Option 2)

7

Incorrect

Option 3)

8

Correct

Option 4)

10

Incorrect

### Preparation Products

##### Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-