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IIT-JEE 2009

PROGRESSION (A.P., G.P., H.P.) (PART-I)

Author: anuraagchandra

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- $a_1, a_2, a_3 \cdots \cdots \cdots a_{n-1}$ , an are said to be in A.P.. If and only if $a_2-a_1=a_3-a_2=\cdots \cdots a_n-a_{n-1}=constant=d(c.d)$
- $a_1, a_2, a_3, \cdots \cdots, a_{n-1}$ , an are in G.P. if and only if $\frac {a_2}{a_1}=\frac {a_3}{a_2}=\cdots \cdots \frac {a_n}{a_{n-1}}=constant=r(c.r.)$
- Unequal numbers $a_1, a_2, a_3, \cdots \cdots , a_{n-1}, a_n$ are H.P. if and only if $\frac {1}{a_1}, \frac {1}{a_2}, \frac {1}{a_3}, \cdots \cdots \frac {1}{a_n}$ are in A.P.
  (This A.P. is called A.P. corresponding to H.P. or simply corresponding A.P.)
  
  Note :
- No term of G.P. or H.P. can be zero.
- Equal non-zero numbers are in A.P., and G.P. both but they are not in H.P.
- Common ratio of a G.P. cannot be zero.
- For A. P.:
- - $t_n(n^{th}term)=a+(n-1)d$
  - $S_n(sum \quad of \quad first \quad n \quad terms)=\frac {n}{2}[first \quad term+last \quad term]$
  - $S_n=\frac {n}{2}[2a+(n-1)d]$
- For G.P.:
- - $t_n=ar^{n-1}$
  - $S_n\frac {a(1-r^n)}{1-r}=\frac {a(r^n-1)}{r-1}$
  - Sum of infinite terms of a G.P., $S_{\infty}=\frac{a}{1-r}$ , where - 1 < r < 1.
  - $n^{th}$ term of an H.P. $= \frac{1}{n^{th} term \quad of \quad corresponding \quad A.P.}$
- c.d. of $A.P.=\frac {b-a}{n-1}$ , where a= first term of A.P., b=last term of A.P. and n=number of term of A.P.
- c.r. of $G.P.=\Big( \frac {b}{a}\Big)^{\frac {1}{n-1}}$ , where a = first term of G.P., b = last term of G.P. and n = number of terms of G.P.
- c.d. of A.P. corresponding to H.P. $\frac {\frac {1}{b}-\frac {1}{a}}{n-1}$ , where a = first term of H.P.,
  b = last term of H.P. and n = number of terms of H.P.
- Arithmetic Mean, A between two numbers a and b is given by A = (a + b)/2 A is the A.M. between a and $b\Leftrightarrow a, A, b$ are in A.P.
- Geometric Mean, G between two positive numbers a an b is given by $G=\sqrt {ab}$ G is the G.M. between two positive numbers a and $b \Rightarrow a, G, b$ are in G.P.
- Harmonic Mean, H between two numbers a and b is given by $H=\frac {2ab}{a+b}$
- $A*H = G^2 \qquad \qquad [ \therefore G^2=ab]$
- A > G > H provided a and b are positive.
- If $A_1,A_2,A_3,\cdots \cdots , A_n$ be n A.M's inserted between a and b, then
- - $a, A_1, A_2, \cdots ,A_n, b$ are in A.P.
  - number of terms of A. P. = n + 2
  - c.d of this $A.P. =\frac {b-a}{n+1}$
  - $k^{th}A.M.=(k+1)^{th}$ term of A.P.
  - sum of n A.M.'s between a and b $=\Big( \frac {a+b}{2}\Big)^n$
  - Explanation:
    $A_1+A_2+ \cdots A_n=\frac {n}{2}(A_1+A_n)$
    $=\frac {n}{2}(a+d+b-d)=\Big( \frac {a+b}{2}\Big) ^n$
- If $G_1,G_2,G_3,\cdots ,G_n$ be n G.M's inserted between a and b, (a, b > 0), then
- - $a, G_1, G_2, \cdots G_n, b$ are in G.P.
  - number of term of G.P. = N + 2
  - c.r. of $G.P.=\Big( \frac {b}{a}\Big) ^{\frac {1}{n+1}}$
  - kth $G.M.=(k+1) ^{th}$ term of G.P.
  - Product of n G.M's between a and $b=(\sqrt {ab}) ^n$
  - Explanation:
    $G_1G_2 \cdots G_n=ar.ar^2.ar^3,\cdots ar^n$
    $=a^nr^{1+2+ \cdots +n}=a^nr$
    $=a^n \Bigg[ \Big( \frac {b}{a} \Big)^{\frac {1}{n+1}}$ $\Bigg] ^{\frac {n(n+1)}{2}}$
    $=a^n\Big(\frac {b}{a}\Big)^{\frac {n}{2}}=a^{\frac {n}{2}}.b^{\frac {n}{2}}=(\sqrt {ab}) ^n$
- If $H_1,H_2,H_3,\cdots ,H_n$ be the n H.M's between a and b, then
  - $a, H_1,H_2,\cdots H_n, b \quad are \quad in \quad H.P.$
  - number of term of H.P. = n + 2
  - c.d. of corresponding A.P. $=\frac {\frac {1}{b}-\frac {1}{a}}{n+1}=\frac {a-b}{(n+1)ab}$
  - $k ^{th}H.M.=(k+1)^{th}$ term of H.P.
If $a_1,a_2,a_3\cdots$ ,an are in A.P. and G.P. both, then
$a_1,=a_2=a_3=\cdots =a_n$
Unequal number cannot be both in A.P. and G.O. at a time.
Explanation:Let be in A.P. and G.P., then

$a_2=\frac {a_1+a_3}{2} \qquad \qquad \cdots (i)$

and $a_2^2 =a_1a_3 \qquad \qquad \cdots (ii)$

$\therefore \frac {(a_1+a_3)^2}{4}=a_1a_3$

$\Rightarrow (a_1 - a_3)^2 = 0$
From

From (i) $a_2=\frac {a_3+a_3}{2}=a_3$

Thus,

Similarly if are in A.P. and G.P., then are in A.P. and G.P.

$\Rightarrow \qquad a_1=a_2=a_3$

Again are in A.P. and G.P.

$\Rightarrow \qquad a_2= a_3= a_4$

Thus
Proceeding in this way, we will get

$a_1=a_2=\cdots \cdots =a_n$