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Question

# The value of x + y + z is 15 if a, x, y, b are in A.P. while the value of 1x+1y+1z is 58 if a, x, y, z b are H.P., then a2+b2=

A

48

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B

50

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C

52

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D

60

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Solution

## The correct option is C 52 a, x, y, b are in A.P ⇒a+b2 x + z = a + d + b – d = a + b, where d is the common difference ∴x+y+z=15⇒a+b2+(a+b)=15⇒a+b=10 a, x, y, z, b are in H.P ⇒1a,1x,1y,1z,1b are in A.P 1x+1y+1z=12(1a+1b)+(1a+1b)=32(1a+1b)=32(a+b)ab=58,given ∴ab=85×32×10=24 [using(1)] Using (1) and (2) ⇒ (a, b) = (4, 6) ⇒a2+b2=52

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