Sequences and Series 5 Question 9

9. (i) The value of x+y+z is 15 . If a,x,y,z,b are in AP while the value of 1x+1y+1z is 53. If a,x,y,z,b are in HP, then find a and b.

(ii) If x,y,z are in HP, then show that log(x+z)+log(x+z2y)=2log(xz).

(1978, 3M)

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Answer:

Correct Answer: 9.(i) a=1,b=9

Solution:

  1. (i) Now, a+b=(a+x+y+z+b)(x+y+z)

=52(a+b)15

[since, a,x,y,z are in AP]

Sum=52(a+b)a+b=10

Since, a,x,y,z,b are in HP, then 1a,1x,1y,1z,1b are in AP.

Now, 1a+1b=1a+1x+1y+1z+1b1x+1y+1z

=521a+1b53

a+bab=109ab=9×1010 [from Eq. (i)]

ab=9

On solving Eqs. (i) and (ii), we get

a=1,b=9

(ii) LHS=log(x+z)+log(x+z2y)

=log(x+z)+logx+z22xzx+zy=2xzx+z=log(x+z)+log(xz)2(x+z)=2log(xz)= RHS 

Topic 6 Relation between AM, GM, HM and Some Special Series



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