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 If x1,x2,xn and 1h1,1h2,.1hn are two A.P's such that x3=h2=8 and x8=h7=20  then x5h10 is 

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a
2560
b
2650
c
3200
d
1600

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detailed solution

Correct option is A

d1 is common difference of A.P of x1,x2…xn then x8−x3=5d1=12⇒d1=2.4  ⇒x5=x3+2d1=8+2.125=12.8 Suppose d2 is the common difference of an A.P 1h1,1h2,⋅⋅1hn1h7-1h2=5d2=120−18=−3/40d2=−3200∵1h10=1h7+3d2=120-9200=1200⇒h10=200∴x5⋅h10=12.8×200=2560


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