Let xn,yn,zn,wn denote nth term of four different arithmetic progressions with positive terms. If x4+y4+z4+w4=8 and x10+y10+z10+w10=20, then maximum possible value of x20⋅y20⋅z20⋅w20 is
A
104
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B
108
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C
1010
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D
1020
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Solution
The correct option is A104 Let xn=ax+(n−1)dx yn=ay+(n−1)dy zn=az+(n−1)dz wn=aw+(n−1)dw
Given x4+y4+z4+w4=8 ⇒ax+ay+az+aw+3(dx+dy+dz+dw)=8⋯(1)
Also x10+y10+z10+w10=20 ⇒ax+ay+az+aw+9(dx+dy+dz+dw)=20⋯(2)
Solving equation (1) and (2). we get dx+dy+dz+dw=2ax+ay+az+aw=2
As the terms of the A.P. is positive, using A.M≥G.M, we get x20+y20+z20+w204≥(x20⋅y20⋅z20⋅w20)1/4⇒ax+ay+az+aw+19(dx+dy+dz+dw)4≥(x20⋅y20⋅z20⋅w20)1/4⇒10≥(x20⋅y20⋅z20⋅w20)1/4⇒x20⋅y20⋅z20⋅w20≤104
Hence, the maximum possible value of the given expression is 104.