If x1,x2,⋯,xn and 1h1,1h2,⋯,1hn are two A.P.s such that x3=h2=8 and x8=h7=20, then x5⋅h10 equals
A
2560
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2650
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3200
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1600
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A2560 We know that, x8=20⇒x1+7d=20x3=8⇒x1+2d=8⇒5d=12⇒x5=x1+4d=x1+2d+2d⇒x5=8+245=645
Now, h7=20⇒1h1+6d=120h2=8⇒1h1+d=18⇒5d=120−18⇒d=−3200⇒1h10=1h1+9d=1h1+6d+3d⇒1h10=120−9200⇒h10=200
Therefore, x5h10=645×200=2560