September 10, 2024

# Sequences and Series Tricks : Arithmetic and Geometric Sequences

Sequences and Series Tricks : Arithmetic and Geometric Sequences

Both arithmetic and geometric sequences  are  important concept of sequences and series . Questions from AP & GP are asked in various competitions including SSC and Railways.  Here, are some important formulas that you can use  to solve questions based on arithmetic and  geometric sequences quickly, easily and efficiently .

Formula 1.    If for an AP., sum of p terms is equal to sum of q terms then sum of (p+q) terms is zero.

Que. 1 If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is   the sum of the first 30 terms?
(1) 0                                               (2) –1
(3) 1                                               (4) Not unique

Sol. (i)   ( proper method)

Sum of 11 terms of an AP equals the sum of 19 terms of the same AP.

$S_{11}=\frac{11}{2}\left&space;[&space;2a+10d&space;\right&space;]$    ……….. (i)          and            $S_{19}=\frac{19}{2}\left&space;[&space;2a+18d&space;\right&space;]$

…………….(ii

On equating  (1) and (2), we get

$\frac{11}{2}\left&space;[&space;2a+10d&space;\right&space;]=\frac{19}{2}\left&space;[&space;2a+18d&space;\right&space;]$

$22a+110d&space;=&space;38a+342d$

0r    $16a&space;=&space;-232d$   . This implies  $a&space;=&space;-(232/16)d$

so   $S_{30}&space;=&space;\frac{30}{2}[2a+29d]$       =$15[-\frac{464d}{16}&space;+&space;29d]$

=  0.

(ii)( By using formula  )

Given that  $S_{11}=S_{19}$  .  So  $S_{30}=S_{11+19}=0$

Formula 2.   If the ratio of sum of ‘$n$‘  terms of two arithmetic progression is  $\frac{f(n)}{g(n)}$

then the ratio of their ‘ $n^{th}$‘   term  will be  $\frac{f(2n-1)}{g(2n-1)}$
.

Que. 2 If the ratio of sum of  $n$ terms of two A.P. is  $\frac{3n+8}{7n+15}$

,  then the ratio of   $12^{th}$   term is  :

(a)  $\frac{16}{7}$

(b)  $\frac{7}{16}$                 (c) $\frac{7}{12}$
(d)  $\frac{12}{5}$

Sol.  (i) (proper method )

Let  $a_{1}$

and  $d_{1}$  be the first term and common difference  of first A.P.  and $a_{2}$
and $d_{2}$ be the first term and common difference of second A.P. , then

$\frac{s_{n}}{s_{n}^{'}}=&space;\frac{3n+8}{7n+15}$

$\Rightarrow&space;\frac{\frac{n}{2}\left&space;[&space;2a_{1}+(n-1)d_{1}&space;\right&space;]}{\frac{n}{2}\left&space;[&space;2a_{2}+(n-1)d_{2}&space;\right&space;]}=\frac{3n+8}{7n+15}$

$\Rightarrow&space;\frac{a_{1}+\frac{(n-1)}{2}d_{1}}{a_{2}+\frac{(n-1)}{2}d_{2}}=\frac{3n+8}{7n+15}$

Now put  $\frac{n-1}{2}=11$. This implies $n=23$

.

So $\frac{a_{1}+11d_{1}}{a_{2}+11d_{2}}=&space;\frac{3\times&space;23+8}{7\times&space;23+15}=\frac{77}{176}=\frac{7}{16}$.

Hence  $\frac{T_{12}}{T_{12}^{'}}=\frac{7}{16}$

(ii)  By using formula:

Given   $\frac{s_{n}}{s_{n}^{'}}=&space;\frac{3n+8}{7n+15}$

replace ‘n’ with 2n-1,we get $\frac{T_{n}}{T_{n}^{'}}=&space;\frac{3(2n-1)+8}{7(2n-1)+15}$

Now put n=12 we get    $\frac{T_{12}}{T_{12}^{'}}=&space;\frac{3\times&space;23+8}{7\times&space;23+15}=\frac{7}{16}$

Formula 3: sum of infinite series of type  $\frac{1}{a_{1}\times&space;b_{1}}+\frac{1}{a_{2}\times&space;b_{2}}+\frac{1}{a_{3}\times&space;b_{3}}+.......................\infty$

=$\frac{1}{a_{1}\times&space;(b_{1}-a_{1})}$ where $a_{1},&space;a_{2},........$
and $\,&space;\,&space;\,&space;\,&space;b_{1},b_{2},.......$    are in A.P.  and $b_{1}-a_{1}=b_{2}-a_{2}=b_{3}-a_{3}=...........$

Que . 3     The sum of the following  series      $\frac{1}{1.4}+&space;\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+............\infty$   is

(a)  $\frac{1}{2}$

(b)  1                 (c) $\frac{1}{3}$                           (d) none

sol.  (i) (proper method )  Let $S_{n}$

be sum of the ‘n’ terms of   the given series , then

$S_{n}=&space;\frac{1}{1.4}+&space;\frac{1}{4.7}+&space;\frac{1}{7.10}+.........+\frac{1}{(3n-2)(3n+1))}$

$=\frac{1}{3}\left&space;[&space;\frac{4-1}{1.4}+\frac{7-4}{4.7}+..........+\frac{(3n+1)-(3n-2)}{(3n+1)(3n-2)}&space;\right&space;]$

=$\frac{1}{3}\left&space;[&space;(\frac{1}{1}-\frac{1}{4})+&space;(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+.................+(\frac{1}{(3n-2)}-\frac{1}{(3n+1)})\right&space;]$

$\Rightarrow&space;S_{n}=\frac{1}{3}\left&space;[&space;1-\frac{1}{3n+1}&space;\right&space;]$

Taking limit  $n\rightarrow&space;\infty&space;,$   we get  $S_{\infty&space;}=&space;lim_{n\rightarrow&space;\infty&space;}\frac{1}{3}\left&space;[&space;1-\frac{1}{3n+1}&space;\right&space;]$

= 3

(ii) (  by using formula)

$\frac{1}{1.4}+&space;\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+............\infty=&space;\frac{1}{1(4-1)}=\frac{1}{3}$

Formula 4:  If  arithmetic mean and geometric mean of two numbers  ‘a’ and ‘b’  (a>b) are in ratio m:n , then  $\frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$

Que.4  Let  $'a'$   and   $'b'$

be positive real numbers such that $a>&space;b$  and $a+b=4\sqrt{ab}$
, then $\frac{a}{b}$  is equal to  (a)  $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
(b) $\frac{3+\sqrt{2}}{3-\sqrt{2}}$       (c) $\frac{4+\sqrt{3}}{4-\sqrt{3}}$
(d) $\frac{3}{2}$

Sol.  (proper  method)   Given that  $a+b=4\sqrt{ab}$

$\Rightarrow\frac{a+b}{2}=2\sqrt{ab}$  which gives

$\frac{\frac{a+b}{2}}{\sqrt{ab}}=\frac{2}{1}$

By applying componendo and dividendo  $\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{2+1}{2-1}$ =$\frac{3}{1}$

$\Rightarrow&space;\frac{(\sqrt{a}+\sqrt{b})^{2}}{(\sqrt{a}-\sqrt{b})^{2}}=\frac{3}{1}$

$\Rightarrow&space;\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}$

By applying componendo and dividend again, we get $\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}+\sqrt{b})}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\Rightarrow&space;\frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\Rightarrow&space;\frac{a}{b}=\left&space;(\frac{\sqrt{3}+1}{\sqrt{3}-1}&space;\right&space;)^{2}$

$\Rightarrow&space;\frac{a}{b}=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}$

=    $\frac{2+\sqrt{3}}{2-\sqrt{3}}$. So the option  (a) is correct .

(ii) (by using formula )  Given that      $\frac{arithmetic\,&space;\,&space;mean&space;}{geometric&space;\,&space;\,&space;mean&space;}=\frac{2}{1}$

.

so $m=2$   and $n=1$

.  Now by using formula   $\frac{a}{b}=\frac{2+&space;\sqrt{2^{2}-1}}{2-\sqrt{2^{2}-1}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

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