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If the sum of the first ten terms of the series `(1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+. . . . . ,` is `(16)/5` m, then m is equal to: (1) 102 (2) 101 (3) 100 (4) 99

A

100

B

99

C

102

D

101

Text Solution

Verified by Experts

The correct Answer is:
D
`((8)/(5))^(2)+((12)/(5))^(2)+((16)/(5))^(2)+((20)/(5))^(2)+"......."+((44)/(5))^(2)`
`=(16)/(25)(2^(2)+3^(2)+4^(2)+5^(2)+"........"+11^(2))`
`=(16)/(25)((11*(11+1)*(22+1))/(6)-1)`
`=(16)/(25)xx505=(16)/(25)xx101=(16)/(5)m(" given ")`
`:.m =101`.
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ARIHANT MATHS-SEQUENCES AND SERIES-Exercise (Questions Asked In Previous 13 Years Exam)
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  13. Let a,b,c be positive integers such that (b)/(a) is an integer. If a,b...

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  15. If m is the AM of two distinct real numbers l and n (l,ngt1) and G(1)...

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  16. Soppose that all the terms of an arithmetic progression (AP) are natur...

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