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Question

If x=112+132+152+...., y=112+322+132+342+.. and z=112122+132142+...., then

A
x,y,z are in A.P.
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B
y6,x3,z1 are in A.P.
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C
y6,x3,z2 are in A.P.
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D
6y,3x,2z are in H.P.
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Solution

The correct option is C y6,x3,z2 are in A.P.

yx=[112+322+132+342+...][112+132+152+...]

=3[122+142+162+...]......(1)

Similarly,

xz=[112+132+152+....][112122+132142+....]

=122+142+162........(2)


From (1) and (2),

yx=3(xz)y+3z=4x

(y6)+(z2)=2(x3) (dividing both sides by 6)

Therefore y6,x3andz2 are in A.P


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