IIT JEE , AIEEE Forum - Mathematics , Algebra

aka_anu

1.calculate the least no. of terms of G.P.5+10+15............whose sum wd exceed1000000

2.sum the series to infinity-

1+1/2-1/4+1/8.....................

3.calculate lim(n tends to infinity){1/2+1+3/2+.........+n/2}/n^2/4+n+3

joyfrancis

q3).the given expression is

1/2 + 2(1/2) + 3(1/2) + 4(1/2)....n(1/2) / {n²/4} + n + 3

=1/2(1 + 2 + 3 + .....n ) / {n²/4} + n + 3

=1/2[n(n+1) / 2] / {n²/4} + n + 3

=(n² + n) / n² + 4n + 12

=( 1 + 1/n ) / ( 1 + {4/n} + {12/n²} )

put n = inf. you will get limit as 1

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nick

let
S =1+1/2-1/4+1/8...............(a)

2S=2+1-1/2+1/4-1/8...............(b)

(a) +(b)
3S=2+1+1 rest of the terms will get cut,as it is till INFINITY
NO TERMS SHALL REMAIN UNCUT as infinity is
undefined

therefore
S=4/3
which i think should be the ans.......

aka_anu

sorry the correct ans is 2

kghedriu

A1) let the last term be 5x...

.: 5+10+....+5x = 1000000

so, 1+2+...+x= 200000

x(x+1)/2 = 200000

x^2+x- 400000=0

hence solve for x..and the requir no. is 5x....

A2) 1+1/2-1/4+1/8......

= 1+ 1/2 { 1-1/2+1/4..}

= 1+ 1/2 { 1/1-1/2}

= 1+ 1/2*2

= 2

A3) the expression under limit is: (1/2+1+3/2+....+n/2)/n^2/4+n+3

= 1/2{1+2+3+....+n}/n^2/4 +n+3

= {1/2*n/2*(n+1)}/ n^2/4+n+3

= {n^2+n}/{n^2+4n+2}

= {n+1/n}/{n+4/n+12/n^2}

as n tends to infinity, 1/n and 1/n^2 tends to 0...

hence answer is 1...

...Goutham Harsha.K

I find bliss in ignorance [^_^]

lazycol

Q1:

its not a gp,but ap

sum n/2( 2a+(n-1)d) = n/2[10+(n-1)5] > 1000000

n(n+1) > 400000

400000 is in bw 321 & 322

so n=632 {as632*633 = 40056}

lazycol

Q2:

sum = 1+1/2 -1/4 +...

= 1 + an infinite gp with a=1/2,r=-1/2

= 1+a/(1-r) = 1+1/2/(1+1/2) = 1+1/3 = 4/3

aansh_c

hy gys/gals 1st of all dnt go 4 d answers

especially mr.kghedriu if dts wt u did

in answer 2

u tuk half common rt

den tuk 1-1/2+1/4 as a GP

wid a common ratio -1/2

& sum 2 infinity is a/1-r

so it shud b

1+ 1/2{1/1+1/2}

1+ 1/2*(2/3)

1+1/3

4/3

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