1.

Insert six A.M.s between 15 and -13.

Answer» Let ` A_(1) , A_(2),A_(3),A_(4),A_(5),A_(6)` be the six arithmetifc means between 15 and -13. Then ,
` 15 , A_(1),A_(2),A_(3),A_(4),A_(5),A_(6),-13` are in AP.
Now , ` d = (( -13 -15))/ (( 6+1)) = (-28)/7 = -4 " " [ because d= ( (b-a))/((n+1)) = ((-13-15))/(( 6+1))]`
` A_(1)= (15+d) = ( 15-4) = 11 , A_(2) = (15+2d) = ( 15-8) =7`,
` A_(3)= (15+3d) = ( 15-12) =3, A_(4) = ( 15 + 4d) = (15 -16) =-1`
` A_(5) = ( 15 +5d) = ( 15-20) = -5, A_(6) = ( 15+6d0 = ( 15-24) =-9`
Hence, the required six AMs between 15 and -13 are
11,7,2,-5 and -9


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