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Question

The value of x+y+z=15, if a,x,y,z,b are in A.P, while the value of 1x+1y+1z=53 if a,x,y,z,b are in H.P. Find a× b.

Solution
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To find the value of a and b

Given,

x+y+z=15....(i)

1x+1y+1z=53........(ii)

if a,x,y,z are in A.P

Then, x,y,z are in A.P

2y=x+z......(iii)

2y+y=15........... comparing (iii) and (i)

3y=15y=5

Now,

a+b2=ya+b=2y

a+b=10......(vi)

Now, if a,x,y,z are in HP

then x,y,z one in H.P

2y=1x+1z.......(iv)

1x+1y+1z=53.......(v)

from (iv) and (v) we can write

2y+1y=53=53........putting (iv) in (v)

3y=53y=95



Also, a,y,b are in H.P.

1a+1b=2y

a+bab=2y=2×59

a+bab=109......(vi)

by comparing (vi) and (vii) we can say that

a+bab=109

10ab=109

ab=9.......(viii)

a(10a)=9

10aa2=9a210a+9=0

(a1)(a9)=0

a=1 or 9

when a=1,b=9

a=9,b=1

a×b=9×1=1×9=9

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