insurt am between 1/2 and 7/4 such that 5/4 is the third a. m., also find the no. of means
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Answer:
8th mean = 9 th term in A.P =t
9
=3+8d
(n-2)th mean = (n-1)th term in A.P =t
(n−1)
=3+(n−2)d
It is given that,
t
n−1
t
9
=
5
3
3+(n−2)d
3+8d
=
5
3
9+3nd−6d=15+40d
3nd=46d+6 or nd=
3
46
d+2..........(1)
If there are n arithmetic means between 3 and 54, then (n+2) terms are there in A.P.
hence, t
n+2
=3+(n+1)d=54 or (n+1)d=nd+d=51 .......................(2)
By substituting nd from eqn.(1) in eqn(2), n is eliminated
3
46
d+2+d=51
⇒46d+6+3d=153
⇒49d=147
and we get d=3.
putting d=3 in (2)
n(3)+3=51
3n=48
Hence n=16
Step-by-step explanation:
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