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Question

# If x=112+132+152+...., y=112+322+132+342+.. and z=112âˆ’122+132âˆ’142+...., 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.y−x=[112+322+132+342+...]−[112+132+152+...]=3[122+142+162+...]......(1)Similarly,x−z=[112+132+152+....]−[112−122+132−142+....]=122+142+162........(2)From (1) and (2),y−x=3(x−z)⟹y+3z=4x⇒(y6)+(z2)=2(x3) (dividing both sides by 6)Therefore y6,x3andz2 are in A.P

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