Question

1) If A. M., G. M. and H. M. of first and last terms of the series 100, 101, 102,.... n - 1, n are the terms of the series itself then the value of n is (100 < n ≤ 500) :
A) 200 B) 300 C) 400 D) 500

2) The sum of the following series to n terms is:

$\sf \dfrac{2}{1.2} + \dfrac{5}{2.3} .2 + \dfrac{10}{3.4} . {2}^{2} + \dfrac{17}{4.5} . {2}^{3} ... \\ \\ \sf \: A) \: \dfrac{n}{n + 1}. {2}^{n + 1} \\ \\ \sf B) \: \dfrac{n + 1}{n} . {2}^{n + 1} \\ \\ \sf C) \: \dfrac{n}{n + 1}. {2}^{n} \\ \\ \sf D) \: \dfrac{n + 1}{n} . {2}^{n}$
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Answer 1

$\huge \text{Question -1 }$

1) If A. M., G. M. and H. M. of first and last terms of the series 100, 101, 102,.... n - 1, n are the terms of the series itself then the value of n is

(100 < n ≤ 500) :

• A) 200

• B) 300

• C) 400

• D) 500

$\bf{Explanation}$

$\sf \implies{A.M} = \frac{100 + n}{2}$

$\sf \implies{G.M} = 10 \sqrt{n}$

$\sf \implies{H.M} = \frac{200n}{100 + n}$

$\bf{Note}$

• For A.M to be integers ,n must be even

• For G.M to be integers ,n must be perfect square

$\bf{So}$

$\sf \implies{n = 4 {k}^{2} }$

$\sf \implies{H.M = \frac{200n}{100 + n} = \frac{200 \times 4 {k}^{2} }{100 + 4 {k}^{2} } = \frac{200 \: {k}^{2} }{25 + {k}^{2} } }$

$\bf{Note}$

• For H.M to be integers k² must be divisible by 25 and also ( 100 < n ≤ 500) :

$\sf \implies {k}^{2} = 50,75,100,125$

$\bf \red{Therefore}$

$\bf \implies{n = 400}$

$\bf{Hence}$

$\bf \implies \boxed{c) \: n = 400} \: is \: correct \: option$

============================

$\huge \text{Question -2}$

2) The sum of the following series to n terms is:

$\sf \dfrac{2}{1.2} + \dfrac{5}{2.3} .2 + \dfrac{10}{3.4} . {2}^{2} + \dfrac{17}{4.5} . {2}^{3} ... \\ \\ \sf \: A) \: \dfrac{n}{n + 1}. {2}^{n + 1} \\ \\ \sf B) \: \dfrac{n + 1}{n} . {2}^{n + 1} \\ \\ \sf C) \: \dfrac{n}{n + 1}. {2}^{n} \\ \\ \sf D) \: \dfrac{n + 1}{n} . {2}^{n}$

$\bf{Explanation}$

$\bf \implies{The \: given \: series \: is }$

$\sf \implies \frac{n}{ \frac{Σ}{r = 1} } = \frac{ {r}^{2} + 1 }{r(r + 1)} \: {2}^{r - 1}$

$\sf \implies \frac{n}{ \frac{Σ}{r = 1} }( \frac{ r }{r + 1} \: {2}^{r} - \frac{r - 1}{r} \: {2}^{r - 1})$

$\bf \red{Therefore}$

$\bf \implies \: C) \: \dfrac{n}{n + 1}. {2}^{n}$

$\bf{Hence}$

$\bf \implies \boxed{ C) \: \dfrac{n}{n + 1}. {2}^{n} } \: is \: correct \: option$

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Answer 2

Answer:

${ \underline{ \boxed{\mathbb{\red{answer \: in \: attachement}}}}}$

Step-by-step explanation:

${ \underline{ \boxed{\mathbb{\purple{hope \: it \: helpful \: for \: you}}}}}$