Which relation is true for A ?

Question

If A = #1/(sqrt1 + sqrt3)# + #1/(sqrt3 + sqrt5)# + #1/(sqrt5 + sqrt7)# + ...... upto 50 terms, then which of the following relation is true for A :

4 #<=# A #<=# 4.5

9 #<=# A #<=# 10

4.5 #<=# A#<=# 5

None of the above

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Answer 1 · 2018-07-20T11:27:36

#"The right option is "(3) 4.5 le A le 5#.

#A=1/(sqrt3+sqrt1)+1/(sqrt5+sqrt3)+1/(sqrt7+sqrt5)+...50"terms"#.

If #A_n# is the #n^(th)# term, then, #A_n=1/(sqrt(2n+1)+sqrt(2n-1))#,

#={sqrt(2n+1)-sqrt(2n-1)}/{(2n+1)-(2n-1)}#.

# rArr A_n=1/2{sqrt(2n+1)-sqrt(2n-1)}#.

#:. A=A_1+A_2+A_3+...+A_49+A_50#,

#=1/2{(cancelsqrt3-sqrt1)+(cancelsqrt5-cancelsqrt3)+(cancelsqrt7-cancelsqrt5)+...#

#+(cancelsqrt99-cancelsqrt97)+(sqrt101-cancelsqrt99)}#.

# :. 2A=sqrt101-1............(ast)#.

Now, #100 lt 101 lt 121#.

#:. sqrt100 lt sqrt101 lt sqrt121,#.

# or, 10 lt sqrt101 lt 11#.

#"Adding" -1," we get, "9 lt sqrt101-1 lt 10#.

#:. (ast ) rArr 9 lt 2A lt 10#.

#"Dividing by "2 gt 0," we get, "4.5 lt A lt 5#.

Hence, the right option is #(3) 4.5 le A le 5#.