Progression & Series

Q: If x = 1 + a + a² + a³ + ….. to ∞ (|a| < 1), y = 1 + b + b² + b³ + ….. to ∞ (|b| < 1) ,  prove that 1 + ab + a²b² + a³b³ + …. to ∞ $ = \frac{x y}{x + y -1}$

Solution : x = 1 + a + a² + a³ + ….. to ∞

$\displaystyle x = \frac{1}{1-a}$

y = 1 + b + b² + b³ + ….. to ∞

$\displaystyle y = \frac{1}{1-b}$

1 + ab + a²b² + a³b³ + …. to ∞

$\displaystyle = \frac{1}{1-a b}$

Putting the values of x and y we get

$\displaystyle \frac{x y}{x + y -1} = \frac{\frac{1}{1-a} . \frac{1}{1-b}}{\frac{1}{1-a} + \frac{1}{1-b} -1}$

$\displaystyle \frac{x y}{x + y -1} = \frac{1}{1-a b} $

Q: Prove that the sum of the n arithmetic means inserted between two quantities is n times the single arithmetic mean between them.

Solution : $\displaystyle AM_r = a + r (\frac{b-a}{n+1})$

$\displaystyle \Sigma AM_r = \Sigma a + \Sigma r (\frac{b-a}{n+1})$

$\displaystyle \Sigma AM_r = n a + \frac{n(b-a)}{2} $

$\displaystyle \Sigma AM_r = n \frac{(a + b)}{2} $

which is n times the single AM between a and b.

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 $

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

Q: Does there exist a G.P. containing 27, 8, and 12 as three of its terms ? If it exists, how many such progressions are possible.

Sol. Let 8 be the m ^th , 12 the n ^th and 27 be the t ^th terms of a G.P. whose first term is A and common ratio is R.

Then , $\displaystyle 8 = AR^{m-1}$ ….(i)

$\displaystyle 12 = AR^{n-1}$ …(ii)

$\displaystyle 27 = AR^{t-1}$ ….(iii)

$\displaystyle \frac{8}{12} = R^{m-n}$

$\displaystyle \frac{12}{27} = R^{n-t}$

$\displaystyle \frac{8}{27} = R^{m-t}$

⇒ 2m – 2n = n – t

and 3m – 3n = m – t

⇒ 2m + t = 3n

and 2m + t = 3n

⇒ $\displaystyle \frac{2m+t}{3} = n $

There are infinity of sets of values of m,n,t which satisfy this relation. For example, take m = 1, then $\displaystyle \frac{2+t}{3} = n = k $ ⇒ n = k , t = 3k-t . By giving different values to k we get integral values of n and t. Hence there are infinite number of G.P.’s whose terms may be 27, 8, 12 (not consecutive).

Q: Suppose that a₁ , a₂, … a_n, a_n+1, … are in A.P. and s_k = a_(k-1)n+1 + a_(k-1)n+2 + … a_kn. Prove that s₁, s₂, s₃, …. are A.P. with the common difference n₂ times the common difference. of the A.P. a₁ , a₂ , a₃……………

Sol. Suppose that a₁, a₂, a₃, …. a_n, a_n+1, …. are in A.P. and

Let d be the c.d. of a₁, a₂ , a₃

s_k = a_(k-1)n+1 + a_(k-1)n+2 + … + a_(k-1)n+n

$\displaystyle = \frac{n}{2}[a_{(k-1)n+1} + a_{(k-1)n+n}] $

$\displaystyle = \frac{n}{2}[ a_1 + ((k-1)n+1 -1)d + a_1+(kn-1)d] $

$\displaystyle = \frac{n}{2}[2a_1 + (2kn -n-1)d ]$

Similarly, $\displaystyle s_{k+1} = \frac{n}{2}[2a_1 + (2kn + n-1)d ] $

s_k+1 – s_k = n²d

s₁, s₂, s₃, … are in A.P. with common difference n² times the common difference of a1, a2, a3, ….

Q: a₁ , a₂ , …. a_n be in arithmetical progression, show that

$\displaystyle a_1^2 a_2^2 ……a_n^2 > a_1^n a_n^n$

Sol. Let d be the common difference of the A.P.

a_r a_{n – r + 1} = {a₁ + (r – 1)d}.{a₁ + (n – r)d}

= a₁² + (n-1)d a₁ + (r-1)(n-r)d²

≥ a₁² + (n-1)d a₁

= a₁[a₁ + (n – 1)d]

a_r.a_{n – r + 1} ≥ a₁ . a_n

a₁ . a_n = a₁ . a_n

a₂ . a_{n – 1} > a₁ . a_n

a₃ . a_{n – 2} > a₁ . a_n

– – – – – – – – – – – – –
– – – – – – – – – – – – –

a_{n – 1} . a₂ > a₁ . a_n

a_n . a₁ = a₁ . a_n

multiplying we get

$\displaystyle a_1^2 . a_2^2 ……a_n^2 > (a_1 . a_n)^n$

Q: If |x| < 1 and |y| < 1 then prove that :

$\displaystyle (x + y) + (x^2 + xy + y^2 ) + (x^3 + x^2 y + xy^2 + y^3 ) + ….. \infty = \frac{x+y-xy}{(1-x)(1-y)} $

Solution : $\displaystyle (x + y) + (x^2 + xy + y^2 ) + (x^3 + x^2 y + xy^2 + y^3 ) + ….. \infty $

Multiplying & dividing by (x-y)

$\displaystyle = \frac{1}{x-y} [(x^2 – y^2) + (x^3 – y^3) + (x^4 – y^4) + …] $

$\displaystyle = \frac{1}{x-y} [(x^2 + x^3 + ….) -(y^2 + y^3 + …) ] $

$\displaystyle = \frac{1}{x-y} [\frac{x^2}{1-x} – \frac{y^2}{1-y} ] $

$\displaystyle = \frac{1}{x-y} (\frac{x^2 -y^2 – x^2y + xy^2 }{(1-x)(1-y)}) $

$\displaystyle = \frac{x+y-xy}{(1-x)(1-y)} $