Question

[
begin{array}{c}
frac{3+8 d}{3+(n-2) d}=frac{3}{5}
9+3 n d-6 d=15+400
3 n d=46 d+6(0)
n d=frac{46}{3} d+2 quad-(1)
end{array}
]
If there are ( n ) A.M b/w sand 54 , then ( (n+2) ) terme are there in
[
A cdot P
]
hence, ( t_{n+2}=3+(n+1) d=54 )
( (a)(n+1) d=n d+d=51-(2) )
By sut nd flom (1) in (2) we get ( d=3 )
[
therefore n=16
]

# There are n A.M's betwee earen A.M's between 3 and 54 such that the 8th mean: (n - 2)th mean:: 3:5. The value of n is. (A) 12 (B) 16 (C) 18 (D) 20 HD

Solution