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KEAM 2011 Math Paper

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Question : 91 of 120

Marks: +1, -0

Let

s_1,s_2,\dots,s_n

be consecutive terms of an A.P. If

\frac{1}{s_1 s_2}+\frac{1}{s_2 s_1}+\dots+\frac{1}{s_{100} s_{101}}=\frac{1}{6}

s_1+s_{101}=50,

|s_1-s_{101}|

is equal to

10
20
30
40
50

Solution:

Let

S_{1}=a, S_{2}=a+d, S_{3}=a+2 d \dots

S_{101}=a+100 d

are in

\text{AP}

Given,

\;\frac{1}{S_{1} S_{2}}+\;\frac{1}{S_{2} S_{3}}+\dots+\;\frac{1}{S_{100} S_{101}}=\;\frac{1}{6}

\Rightarrow \frac{1}{a(a+d)} +\;\frac{1}{(a+d)(a+2 d)}+\dots+\;\frac{1}{(a+99 d)(a+100 d)}=\;\frac{1}{6}

\Rightarrow \;\frac{1}{d}\left\{\;\frac{1}{a}-\;\frac{1}{a+d}\right\}+\;\frac{1}{d}\left\{\;\frac{1}{a+d}-\;\frac{1}{a+2 d}\right\}+\dots+\;\frac{1}{d}\left\{\;\frac{1}{a+99 d}-\;\frac{1}{a+100 d}\right\}=\;\frac{1}{6}

\Rightarrow \;\; \;\frac{1}{a}-\;\frac{1}{a+100 d}=\;\frac{d}{6}

\Rightarrow \;\; \;\frac{100 d}{a(a+100 d)}=\;\frac{d}{6} \Rightarrow 600=a(a+100 d)

\Rightarrow \;\; S_{101}=\;\frac{600}{a}

....(i)

Given that,

\;\; S_{1}+S_{101}=50

a+\;\frac{600}{a}=50

\Rightarrow \;\; a^{2}-50 a+600=0

\Rightarrow \;\; a^{2}-30 a-20 a+600=0

\Rightarrow \;\; a(a-30)-20(a-30)=0

\Rightarrow \;\; (a-30)(a-20)=0

\Rightarrow \;\; a=20,30

So,

\;\; S_{1}=20

or

S_{1}=30

and

\;\; S_{101}=a+100 d=\;\frac{600}{a}

[from Eq. (i)]

When

a=20,20+100 d=\;\frac{600}{20}=30

100 d=10 \Rightarrow d=\;\frac{1}{10}

When

a=30,30+100 d=\;\frac{600}{30}=20

d=-\;\frac{1}{10}

So, when

S_{1}=20,

then,

S_{101}=a+100 d

=20+\;\frac{100}{10}=30

when

S_{1}=30,

then

S_{101}=a+100 d

=30-\;\frac{100}{10}=20

Hence,

(i) When

S_{1}=20, S_{101}=30

|S_{1}-S_{101}|=|20-30|=10

(ii) When

S_{1}=30, S_{101}=20

|S_{1}-S_{101}|=|30-20|=10

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