1.

There are m arithmetic means between 5 and -16 such that the ratio of the 7th mean to the (m-7)th mean is 1:4. Find the value of m.

Answer»

Solution :LET `A_(1),A_(2),...,A_(m)`be m arithmetic means between 5 and -16.
`:. 5,A_(1),A_(2),A_(3),...A_(m)-16` are in A.P.
`:. T_(m+2)=-16 RARR 5+(m+1)d=-16`
`rArr d=(-21)/(m+1)`
`rArr (A_(7))/(A_(m-7))=(1)/(4) rArr(T_(8))/(T_(m-6))=(1)/(4)`
`rArr (5+7d)/(5+(m-7)d)=(1)/(4) rArr (5+7((-21)/(m+1)))/ (5+(m-7)((-21)/(m+1)))=(1)/(4)`
`rArr 20-(588)/(m+1)=5-(21(m-7))/(m+7)`
`rArr 20m+20-588=5m+5-21m+147`
`rArr36m=720`
`rArr m=20`


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