Progression & Series

Q: If a, b, x, y are positive natural numbers such that $\displaystyle \frac{1}{x} + \frac{1}{y} = 1$ then Prove that $\displaystyle \frac{a^x}{x} + \frac{a^y}{y} = \ge a b $

Sol: Consider the positive numbers

a^x , a^x , …. ky times and b^y, b^y, …. kx times

For all these numbers,

$\displaystyle \frac{(a^x + a^x + …..ky \; times )+(b^y + b^y + …..k x \; times)}{k x + k y} $

$\displaystyle = \frac{ky a^x + k x b^y}{k(x+y)} $

$\displaystyle = \frac{y a^x + x b^y}{(x+y)} $

$\displaystyle GM = [(a^x .a^x …..ky \; times) (b^y . b^y ……kx \times)]^{1/k(x+y)} $

$\displaystyle GM =( a^{x(ky) . b^{y(kx)}})^{1/k(x+y)} $

$\displaystyle GM = (ab)^{(\frac{k xy}{k(x+y)})} = (ab)^{(\frac{xy}{(x+y)})} $ …..(i)

As , $\displaystyle \frac{1}{x} + \frac{1}{y} = 1 $

$\displaystyle \frac{x+y}{x y} = 1 $

x + y = x y

(i) becomes

$\displaystyle \frac{y a^x + x b^y}{x y} \ge a b $

$\displaystyle \frac{a^x}{x} + \frac{b^y}{y} \ge a b $

Q: If a1, a2, a3 ……..an are positive and (n – 1)s = a1 + a2 + …… + an then prove that a1 a2 a3 ….an ≥ (n – 1)n (s – a1)(s – a2)……(s – an)

Sol: $\displaystyle \frac{(s-a_2) + (s-a_3) + ….+ (s-a_n)}{n-1} \ge [(s-a_2) (s-a_3) …..(s-a_n)]^{1/n-1}$

$\displaystyle \frac{a_1}{n-1} \ge [(s-a_2) (s-a_3) …..(s-a_n)]^{1/n-1}$

$\displaystyle \frac{a_2}{n-1} \ge [(s-a_1) (s-a_3) …..(s-a_n)]^{1/n-1}$

$\displaystyle \frac{a_3}{n-1} \ge [(s-a_1) (s-a_2)(s-a_4) …..(s-a_n)]^{1/n-1}$

– – – – – – – – – –

– – – – – – – – –

$\displaystyle \frac{a_n}{n-1} \ge [(s-a_1) (s-a_2) …..(s-a_{n-1})]^{1/n-1}$

multiplying, we get

$\displaystyle \frac{a_1 . a_2 . a_3 ….a_n}{(n-1)^n} \ge [(s-a_1) (s-a_2)(s-a_4) …..(s-a_n)] $

$\displaystyle a_1 . a_2 . a_3 ….a_n \ge (n-1)^n[(s-a_1) (s-a_2)(s-a_4) …..(s-a_n)] $

Q: If the sum of m terms of an arithmetical progression is equal to the sum of either the next n terms or the next p terms, prove that $\displaystyle (m+n)(\frac{1}{m}-\frac{1}{p}) = (m+p)(\frac{1}{m}-\frac{1}{n})$

Sol. Let first term = a Common difference = d,

Sm = sum of first m terms.

Then given S_m = S_{m + n} – S_m = S_{m + p} – S_m

S_m = S_{m + n} –S_m ⇒ 2S_m = S_{m + n}

$\displaystyle 2 \frac{m}{2}(2a + (m-1)d) = \frac{m+n}{2}(2a+(m+n-1)d)$

2a [2m – m – n] = d [m² + n² + 2mn – m – n – 2m² + 2m]

2a [m – n] = d [ n² – m² + 2mn + m – n ] ….(i)

also 2 S_m = S_{m + p}

2a [m – p] = d [p² – m² + 2mp + m – p] ….(ii)

from (i) and (ii)

$\displaystyle \frac{n^2 – m^2 + 2 m n + m – n}{m-n} = \frac{p^2 -m^2 + 2 m p + m-p}{m-p}$

$\displaystyle -(m+n) + 1 + \frac{2 mn}{m-n} = -(m+p) + 1 + \frac{2 m p}{m-p} $

$\displaystyle \frac{2 mn – m n + n^2}{m-n} = \frac{2 m p – m p + p^2}{m-p} $

$\displaystyle \frac{(m+n)n}{m-n} = \frac{(m+p)p}{m-p} $

$\displaystyle (m+n)(\frac{m-p}{m p}) = (m+p)(\frac{m-n}{m n})$

$\displaystyle (m+n)(\frac{1}{p}-\frac{1}{m}) = (m+p)(\frac{1}{n}-\frac{1}{m})$

$\displaystyle (m+n)(\frac{1}{m}-\frac{1}{p}) = (m+p)(\frac{1}{m}-\frac{1}{n})$

Q: Let $\displaystyle S= \Sigma_{i=1}^{n} a_i $ and Show that $\displaystyle \Sigma_{i=1}^{n} \frac{s}{s-a_i} > \frac{n^2}{n-1}$ assuming not all a_i ‘s are equal.

Sol: $\displaystyle AM_1 = \frac{1}{n} \Sigma_{i=1}^{n} \frac{s}{s-a_i} $

$\displaystyle GM_1 = ( \Pi_{i=1}^{n} \frac{s}{s-a_i})^{1/n} $ ;( ∏ denotes product sign )

And AM₁ > GM₁ (as a_i ‘s are not all equal) ….. (i)

Again , $\displaystyle AM_2 = \frac{1}{n} \Sigma_{i=1}^{n} \frac{s-a_i}{s} $

$\displaystyle GM_2 = ( \Pi_{i=1}^{n} \frac{s-a_i}{s})^{1/n} $

And AM₂ > GM₂ …. (ii)

By multiplying (i) and (ii)

$\displaystyle \frac{1}{n^2} (\Sigma_{i=1}^{n} \frac{s}{s-a_i}) ( \Sigma_{i=1}^{n} \frac{s-a_i}{s}) > 1 $

$\displaystyle \frac{1}{n^2} (\Sigma_{i=1}^{n} \frac{s}{s-a_i}) (\frac{ns – (a_1 + a_2 + ….+ a_n )}{s} )> 1 $

$\displaystyle \Sigma_{i=1}^{n} \frac{s}{s-a_i} > \frac{n^2}{n-1}$