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 [ 2a+10d \right ]

    ……….. (i)          and            S_{19}=\frac{19}{2}\left [ 2a+18d \right ]    …………….(ii

On equating  (1) and (2), we get

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

22a+110d = 38a+342d    0r    16a = -232d   . This implies  a = -(232/16)d

so   S_{30} = \frac{30}{2}[2a+29d]       =15[-\frac{464d}{16} + 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}^{'}}= \frac{3n+8}{7n+15}

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

\Rightarrow \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}}= \frac{3\times 23+8}{7\times 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}^{'}}= \frac{3n+8}{7n+15}

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

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

Formula 3: sum of infinite series of type  \frac{1}{a_{1}\times b_{1}}+\frac{1}{a_{2}\times b_{2}}+\frac{1}{a_{3}\times b_{3}}+.......................\infty=\frac{1}{a_{1}\times (b_{1}-a_{1})} where a_{1}, a_{2},........     and \, \, \, \, 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}+ \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}= \frac{1}{1.4}+ \frac{1}{4.7}+ \frac{1}{7.10}+.........+\frac{1}{(3n-2)(3n+1))}

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

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

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

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

(ii) (  by using formula)

\frac{1}{1.4}+ \frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+............\infty= \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> 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 \frac{(\sqrt{a}+\sqrt{b})^{2}}{(\sqrt{a}-\sqrt{b})^{2}}=\frac{3}{1}

\Rightarrow \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 \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}

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

\Rightarrow \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\, \, mean }{geometric \, \, mean }=\frac{2}{1}.

so m=2   and n=1 .  Now by using formula   \frac{a}{b}=\frac{2+ \sqrt{2^{2}-1}}{2-\sqrt{2^{2}-1}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}

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