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 $\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}+\sqrt{b})}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$