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Sequences and Series question

2018 · 16 Apr · Shift 1 · Q31
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Sequences and Series question

2018 · 16 Apr · Shift 1 · Q31

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let 1x1,1x2,...,1xn  {1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,x1​1​,x2​1​,...,xn​1​(xi eee 0 for i = 1, 2, ..., n) be in A.P. such that x1=4 and x21 = 20. If n is the least positive integer for which xn>50,{x_n} \gt 50,xn​>50, then ∑i=1n(1xi)\sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)}i=1∑n​(xi​1​) is equal to :
  1. A
    18{1 \over 8}81​
  2. B
    3
  3. C
    138{{13} \over 8}813​
  4. D
    134{{13} \over 4}413​
View written solutionFree

Correct answer: D

  1. Since 1x1,1x2,…,1xn\dfrac{1}{x_1}, \dfrac{1}{x_2}, \dots, \dfrac{1}{x_n}x1​1​,x2​1​,…,xn​1​ are in A.P., let
1xk=a+(k−1)d.\frac{1}{x_k}=a+(k-1)d.xk​1​=a+(k−1)d.

Given x1=4x_1=4x1​=4, so

1x1=14.\frac{1}{x_1}=\frac14.x1​1​=41​.

Hence,

a=14.a=\frac14.a=41​.
  1. Also given x21=20x_{21}=20x21​=20, so
1x21=120.\frac{1}{x_{21}}=\frac{1}{20}.x21​1​=201​.

But

1x21=a+20d=14+20d.\frac{1}{x_{21}}=a+20d=\frac14+20d.x21​1​=a+20d=41​+20d.

Therefore,

14+20d=120.\frac14+20d=\frac1{20}.41​+20d=201​.

So,

20d=120−14=1−520=−420=−1520d=\frac1{20}-\frac14=\frac{1-5}{20}=-\frac4{20}=-\frac1520d=201​−41​=201−5​=−204​=−51​

which gives

d=−1100.d=-\frac{1}{100}.d=−1001​.
  1. Thus the general term is
1xk=14−(k−1)1100=25−(k−1)100=26−k100.\frac{1}{x_k}=\frac14-(k-1)\frac{1}{100} =\frac{25-(k-1)}{100} =\frac{26-k}{100}.xk​1​=41​−(k−1)1001​=10025−(k−1)​=10026−k​.

Hence,

xk=10026−k.x_k=\frac{100}{26-k}.xk​=26−k100​.
  1. We need the least positive integer nnn for which
xn>50.x_n>50.xn​>50.

Using the formula,

10026−n>50.\frac{100}{26-n}>50.26−n100​>50.

Since values up to the required nnn have positive denominator, solve:

100>50(26−n)100>50(26-n)100>50(26−n) 2>26−n2>26-n2>26−n n>24.n>24.n>24.

Therefore the least positive integer is

n=25.n=25.n=25.

Check:

x24=1002=50,x25=1001=100>50.x_{24}=\frac{100}{2}=50, \qquad x_{25}=\frac{100}{1}=100>50.x24​=2100​=50,x25​=1100​=100>50.

So indeed n=25n=25n=25.

  1. Now compute
∑i=1n1xi=∑i=12526−i100.\sum_{i=1}^n \frac{1}{x_i}=\sum_{i=1}^{25} \frac{26-i}{100}.i=1∑n​xi​1​=i=1∑25​10026−i​.

This is

1100∑i=125(26−i).\frac{1}{100}\sum_{i=1}^{25}(26-i).1001​i=1∑25​(26−i).

As iii runs from 111 to 252525, (26−i)(26-i)(26−i) runs from 252525 down to 111. So

∑i=125(26−i)=1+2+⋯+25=25⋅262=325.\sum_{i=1}^{25}(26-i)=1+2+\cdots+25=\frac{25\cdot 26}{2}=325.i=1∑25​(26−i)=1+2+⋯+25=225⋅26​=325.

Hence,

∑i=1251xi=325100=134.\sum_{i=1}^{25}\frac{1}{x_i}=\frac{325}{100}=\frac{13}{4}.i=1∑25​xi​1​=100325​=413​.
  1. Therefore the correct option is
134.\boxed{\frac{13}{4}}.413​​.

This matches option D\boxed{D}D​.

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