Arithmetico-geometric Series

A series is said to be an arithmetico geometric series if its each term is formed by multiplying the corresponding terms of an A.P and a G.P.

E.g. $1+2 \mathrm{x}+3 \mathrm{x}^{2}+4 \mathrm{x}^{3}+$.

Here $1,2,3,4 \ldots$. are in A.P and $1, x, x^{1}, x^{3} \ldots \ldots$ are in G.P

Sum to $n$ terms

Let $S _{n}=a+(a+d) r+(a+2 d) r^{2}+\ldots \ldots .(a+(n-1) d) r^{n-1}………(1)$

Multiply by $\mathrm{r}$ on both the sides

$\begin{aligned} & r S_n=a r+(a+d) r^2+\ldots \ldots . .(a+(n-1) d) r^n \ldots \ldots .(2) \\ & (1)-(2) \Rightarrow S_n(1-r)=a+\left(d r+d r^2+\ldots . . d r^{n-1}\right)-(a+(n-1) d) r^n \\ & (n-1) \text {terms}\end{aligned} $

$\begin{aligned} & a+d r \frac{\left(1-r^{n-1}\right)}{1-r}-(a+(n-1) d) r^n \\ & S_n==\frac{a}{1-r}+\frac{d r\left(1-r^{n-1}\right)}{(1-r)^2}-\frac{(a+(n-1) d) r^n}{1-r}\end{aligned}$

Sum to infinity

If $|\mathrm{r}|<1$ and $\mathrm{n} \rightarrow \infty$, then $\lim _{\mathrm{n} \rightarrow \infty} \mathrm{r}^{\mathrm{n}}=0$

$\therefore \quad \mathrm{S}=\frac{\mathrm{a}}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1-\mathrm{r})^{2}}$

Note: If we take the first term of a G.P to be b, then

$\mathrm{S} _{\mathrm{n}}=\frac{\mathrm{ab}}{1-\mathrm{r}}+\frac{\mathrm{dbr}\left(1-\mathrm{r}^{\mathrm{n}-1}\right)}{(1-\mathrm{r})^{2}}-\frac{(\mathrm{a}+(\mathrm{n}-1) \mathrm{d}) \mathrm{br}^{\mathrm{n}}}{1-\mathrm{r}}$

If $|\mathrm{r}|<1$, then sum to infinity, $S=\frac{\mathrm{ab}}{1-\mathrm{r}}+\frac{\mathrm{dbr}}{(1-\mathrm{r})^{2}}$

Use of Natural numbers

$1 \quad $ Let $\mathrm{S} _{\mathrm{r}}=1^{\mathrm{r}}+2^{\mathrm{r}}+3^{\mathrm{r}}+$ $……….+\mathrm{n}^{\mathrm{r}}$, then

(i) $\mathrm{S} _{1}=1+2+3+\ldots \ldots \ldots \ldots \ldots+\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$

(ii) $\mathrm{S} _{2}=1^{2}+2^{2}+3^{3}+\ldots \ldots \ldots \ldots \ldots \ldots+\ldots \ldots+\mathrm{n}^{2}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$

(iii) $\mathrm{S} _{3}=1^{3}+2^{3}+3^{3}+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{n}^{3}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}}{4}=\mathrm{S} _{1}{ }^{2}$

(iv) $\mathrm{S} _{4}=1^{4}+2^{4}+3^{4}+\ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{n}^{4}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\left(3 \mathrm{n}^{2}+3 \mathrm{n}-1\right)}{30}=\frac{\mathrm{S} _{2}}{5}\left(6 \mathrm{~S} _{1}-1\right)$

(iv) $\mathrm{S} _{5}=1^{5}+2^{5}+3^{5}+\ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{n}^{5}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}\left(2 \mathrm{n}^{2}+2 \mathrm{n}-1\right)}{12}=\frac{1}{3} \mathrm{~S} _{1}{ }^{2}\left(4 \mathrm{~S} _{1}-1\right)$

$2 \quad 1+3+5+$ ……….to $\mathrm{n}$ terms $=\mathrm{n}^{2}$

$3 \quad 1^{2}+3^{2}+5^{2}+$……….to $\mathrm{n}$ terms $=\frac{\mathrm{n}\left(4 \mathrm{n}^{2}-1\right)}{3}$

$4 \quad 1^{3}+3^{3}+5^{3}+$………..to $\mathrm{n}$ terms $=\mathrm{n}^{2}\left(2 \mathrm{n}^{2}-1\right)$

$5 \quad 1-1+1-$ ………..to $\mathrm{n}$ terms $=\frac{1-(-1)^{\mathrm{n}}}{2}$

$6 \quad 1-2+3-$ ………..to $\mathrm{n}$ terms $=\frac{1-(-1)^{\mathrm{n}}(2 \mathrm{n}+1)}{4}$

$7 \quad 1^{2}-2^{2}+3^{2}-$ ………..to $\mathrm{n}$ terms $=\frac{(-1)^{\mathrm{n}-1} \mathrm{n}(\mathrm{n}+1)}{2}=(-1)^{\mathrm{n}-1} \mathrm{~S} _{1}$

$8 \quad 1^{3}-2^{3}+3^{3}-$ ………..to $n$ terms $=\frac{(-1)^{n-1}\left(4 n^{3}+6 n^{2}-1\right)-1}{8}$

Application

If $\mathrm{n}^{\text {th }}$ term of a sequence is given by $\mathrm{T} _{\mathrm{n}}=\mathrm{an}^{3}+\mathrm{bn}^{2}+\mathrm{cn}+\mathrm{d}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{R}$, then

$ \begin{aligned} & \mathrm{S} _{\mathrm{n}}=\sum \mathrm{T} _{\mathrm{n}}=\mathrm{T} _{1}+\mathrm{T} _{2}+\ldots \ldots \ldots \ldots \ldots+\mathrm{T} _{\mathrm{n}} \\ & =\mathrm{a} \sum \mathrm{n}^{3}+\mathrm{b} \sum \mathrm{n}^{2}+\mathrm{c} \sum \mathrm{n}+\mathrm{d} \sum 1 \end{aligned} $

Note: If $\mathrm{T} _{\mathrm{n}}$ is expressible as product of $\mathrm{m}$ consecutive numbers beginning with $\mathrm{n}$, i.e. $T _{n}=n(n+1)(n+2) \ldots(n+m-1)$ then

$ \mathrm{S} _{\mathrm{n}}=\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{m}-1)\left(\frac{\mathrm{n}+\mathrm{m}}{\mathrm{m}+1}\right) $

Eg. If $\mathrm{T} _{\mathrm{n}}=\mathrm{n}$, then $\mathrm{S} _{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$ $\quad ($ Here $\mathrm{m}=1)$

E.g. $T _{n}=n(n+1)$, then $S _{n}=\frac{n(n+1)(n+2)}{3}$ $\quad ($ Here $\mathrm{m}=2)$

Method of differences

If the differences of successive terms of a series are in A.P. or G.P., we can find $T _{n}$ as follows

(a) Denote $\mathrm{n}^{\text {th }}$ term and the sum up to $n$ terms by $T _{n} \& S _{n}$ respectively

(b) Rewrite the given series with each term shifted by one place to the right

(c) Subtracting the above two forms of the series, find $T_{n}$.

(d) Apply $S _{n}=\sum T _{n}$.

Note: Instead of determining the $\mathrm{n}^{\text {th }}$ item of a series by the method of difference, we can use the following steps to obtain the same

(i) If the differences $\mathrm{T} _{2}-\mathrm{T} _{1}, \mathrm{~T} _{3}-\mathrm{T} _{2}$ $\mathrm{T} _{\mathrm{n}}=\mathrm{an}^{2}+\mathrm{bn}+\mathrm{c}, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$

Determine $a, b, c$ by putting $n=1,2,3$ and equating them with the values of corresponding terms of the given series.

(ii) If the differences $T _{2}-T _{1}, T _{3}-T _{2}, \ldots \ldots \ldots \ldots$…tc are in G.P, with common ratio $\mathrm{r}$, then take $T _{n}=$ $a^{n-1}+b n+c, a, b, c \in R$

Determine $\mathrm{a}, \mathrm{b}, \mathrm{c}$ by putting $\mathrm{n}=1,2,3$ and equating them with the values of corresponding terms of the given series.

(iii) If the differences of the differences computed in step (i) are in A.P, then take $T _{n}=a^{3}+$ $\mathrm{bn}^{2}+\mathrm{cn}+\mathrm{d}$

Determine $a, b$, by putting $n=1,2,3,4$ and equating them with the values of corresponding terms of the given series.

(iv) If the differences of differences computed in step (i) are in G.P with common ratio $\mathrm{r}$, then take $\mathrm{T} _{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}+\mathrm{bn}^{2}+\mathrm{cn}+\mathrm{d}$

Determine $a, b$, by putting $n=1,2,3,4$ and equating them with the values of corresponding terms of the given series.

Summation by " $\sum$ " (sigma) operator

i. $ \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T} _{\mathrm{r}}=\mathrm{T} _{1}+\mathrm{T} _{2}+\ldots \ldots .+\mathrm{T} _{\mathrm{n}}$

ii. $ \sum _{\mathrm{r}=1}^{\mathrm{n}} 1=1+1+1+\ldots \ldots+\mathrm{n}$ times $=\mathrm{n}$

iii. $ \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{k} \mathrm{T} _{\mathrm{r}}=\mathrm{k} \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T} _{\mathrm{r}} ; \mathrm{k}$ is a constant

iv. $ \sum _{r=1}^{n}\left(T _{r} \pm T _{r}^{1}\right)=\sum _{r=1}^{n} T _{r} \pm \sum _{r=1}^{n} T _{r}^{1}$

v. $ \sum _{j=1}^{n} \sum _{i=1}^{n} T _{i} T _{j}=\left(\sum _{i=1}^{n} T _{i}\right)\left(\sum _{j=1}^{n} T _{j}\right)$

(Note that $\mathrm{i} \& \mathrm{j}$ are independent here)

vi. Now consider $\sum _{0 \leq i<j \leq n} f _{n}(\mathrm{i}) \times \mathrm{f}(\mathrm{j})$ Here three types of terms occur, for which $\mathrm{i}<\mathrm{j}, \mathrm{i}>\mathrm{j}$ and $i=j$. Also note that the sum of terms when $i<j$ equal to the sum of the terms when $i>j$ if $f(i)$ and $f(j)$ are symmetrical. In such case,

$ \begin{aligned} & \sum _{i=0}^{n} \sum _{j=0}^{n} f(i) f(j)=\sum _{0 \leq j<i \leq n} \sum _{i} f(i) f(j)+\sum _{0 \leq i<j \leq n} \sum _{i=j} f(i) f(j)+\sum _{i} \sum _{f}(i) f(j) \\ & =\quad 2 \sum _{0 \leq i<j \leq n} \sum _{f} f(i) f(j)+\sum _{i=j} \sum f(i) f(j) \\ & \Rightarrow \quad \sum _{0 \leq i<j \leq n} f(i) f(j)=\frac{\sum _{i=0}^{n} \sum _{j=0}^{n} f(i) f(j)-\sum _{i} \sum _{=j} f(i) f(j)}{2} \end{aligned} $

When $f(i)$ and $f(j)$ are not symmetrical, the sum can be obtained by listing all the terms.

Example: 1 $\sum _{\mathrm{i}=1}^{4} \sum _{\mathrm{j}=1}^{3} \mathrm{ij}=\sum \mathrm{i}(1+2+3)=\sum(\mathrm{i}+2 \mathrm{i}+3 \mathrm{i})$

$ (1+2+3)+(2+4+6)+(3+6+9)+(4+8+12)=6+12+18+24=60 $

Also, $\sum _{\mathrm{i}=1}^{4} \sum _{\mathrm{j}=1}^{3} \mathrm{i}=\sum _{\mathrm{i}=1}^{4} \mathrm{i} \sum _{\mathrm{j}=1}^{3} \mathrm{j}=\frac{4 \mathrm{x} 5}{2} \mathrm{x} \frac{3 \times 4}{2}=60 \quad$ (Since $\mathrm{i} \& \mathrm{j}$ are independent)

Note that $2 \sum _{\mathrm{i}<\mathrm{j}=1}^{\mathrm{n}} \mathrm{a} _{\mathrm{i}} \mathrm{a} _{\mathrm{j}}=\left(\mathrm{a} _{1}+\mathrm{a} _{2}+\ldots . .+\mathrm{a} _{\mathrm{n}}\right)^{2}-\left(\mathrm{a} _{1}^{2}+\mathrm{a} _{2}^{2}+\ldots . .+\mathrm{a} _{\mathrm{n}}^{2}\right)$

Example: 2 $ \sum \sum _{0 \leq i \mathrm{j} j \mathrm{n}} 1$

$ \begin{aligned} & =\frac{\sum _{i=1}^{n} \sum _{j=1}^{n} 1-\sum _{i=j} \sum 1}{2} \\ & =\frac{\left(\sum _{i=1}^{n} 1\right)\left(\sum _{j=1}^{n} 1\right)-\sum _{j=1}^{n} 1}{2} \\ & =\quad \frac{n \cdot n-n}{2}=\frac{n(n-1)}{2}={ }^{n} C _{2} \end{aligned} $

Example: 3 Consider $\sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \mathrm{ij}$

There are 3 types of terms in this summation,

i. Those terms when $\mathrm{i}<\mathrm{j}$ (upper triangle)

ii. Those terms when $\mathrm{i}>\mathrm{j}$ (lower triangle)

iii. Those terms when $\mathrm{i}=\mathrm{j}$ (diagonal) It is shown in the diagram

$\sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \mathrm{ij}=$ sum of terms in upper triangle + sum of terms in lower triangle + sum of terms in diagonal.

$ \begin{aligned} & \sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \mathrm{ij}=2 \sum _{0 \leq i \leq j \leq \mathrm{n}} \sum _{\mathrm{i} j} \mathrm{ij}+\sum _{\mathrm{i}=\mathrm{j}} \sum _{\mathrm{ij}}(\because \text { sum of terms in upper }+ \text { lower tringles are same }) \\ & \sum _{0 \leq i \leq j \leq n} \sum _{\mathrm{ij}}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \mathrm{ij}-\sum _{\mathrm{i}=\mathrm{j}} \sum \mathrm{ij}}{2}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \mathrm{i}-\sum _{\mathrm{i}=1}^{\mathrm{n}} \mathrm{i}^{2}}{2} \\ & =\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2} \cdot \frac{\mathrm{n}(\mathrm{n}+1)}{2}-\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}}{2} \\ & =\frac{\mathrm{n}\left(\mathrm{n}^{2}-1\right)(3 \mathrm{n}+2)}{24} \end{aligned} $

Solved examples

1. Find sum of the series to $n$ terms

2. $3^{\frac{1}{4}} \times 9^{\frac{1}{8}} \times 27^{\frac{1}{16}} \times \ldots \ldots .$. to $\infty=$

(a). $3$

(b). $9$

(c). $\frac{1}{3}$

(d). none of these

3. $\sum _{\mathrm{m}=1}^{\infty} \sum _{\mathrm{n}=1}^{\infty} \frac{\mathrm{m}^{2} \mathrm{n}}{3^{\mathrm{m}}\left(\mathrm{n} \cdot 3^{\mathrm{m}}+\mathrm{m} \cdot 3^{\mathrm{n}}\right)}=$

(a). $\frac{16}{9}$

(b). $\frac{9}{16}$

(c). $1$

(d). none of these

4. Sum to $n$ terms of the series $1^{2}-2^{2}+3^{2}-4^{2}+$ is

5. Find the sum of all possible products of first $n$ natural numbers taken two by two

Solution: $\sum _{1 \leq i \leq j \leq n} \mathrm{x} _{\mathrm{i}} \mathrm{x} _{\mathrm{j}}=\frac{1}{2}\left\{\sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \mathrm{ij}-\sum _{\mathrm{i}=\mathrm{j}} \sum \mathrm{ij}\right\}$

$ \begin{aligned} & =\quad \frac{1}{2}\left\{\left(\frac{n(n+1)}{2}\right)^{2}-\frac{n(n+1)(2 n+1)}{6}\right\} \\ & =\quad \frac{1}{24} n(n+1)(n-1)(3 n+2) \end{aligned} $

6. Find sum to $\mathrm{n}$ terms of $\frac{1}{1+1^{2}+1^{4}}+\frac{2}{1+2^{2}+2^{4}}+\frac{3}{1+3^{2}+3^{4}}+\ldots \ldots \ldots$

7. $\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\frac{9}{400}+$ …….to $\infty$ is

(a). 2

(b). 1

(c). 3

(d). none of these

8. Find the $n^{\text {th }}$ term of the following series

i. $3+7+13+21+……$

ii. $5+7+13+31+85+…….$

$V _{n}$ method

1. If $S _{n}=\frac{1}{a _{1} a _{2} \ldots \ldots . a _{r}}+\frac{1}{a _{2} a _{3} \ldots \ldots . a _{r+1}}+\ldots \ldots . .+\frac{1}{a _{n} a _{n+1} a _{n} \ldots \ldots a _{n+r-1}}$

$T _{n}=\frac{1}{a _{n} a _{n+1} \ldots \ldots a _{n+r-1}}, V _{n}=\frac{1}{a _{n+1} \ldots \ldots a _{n+r-1}} \quad$ (leave 1st term in denominator of $T _{n}$ )

$V _{n}-V _{n-1}=\frac{1}{a _{n+1} a _{n+2} \ldots \ldots a _{n+r-1}}-\frac{1}{a _{n} a _{n+1} \ldots \ldots a _{n+r-2}}$

$=\frac{1}{a _{n} a _{n+1} \ldots \ldots a _{n+r-1}}\left(a _{n}-a _{n+r-1}\right)$

$\mathrm{V} _{\mathrm{n}}-\mathrm{V} _{\mathrm{n}-1}=\mathrm{T} _{\mathrm{n}}(1-\mathrm{r}) \mathrm{d}$

$\mathrm{T} _{\mathrm{n}}=\frac{-1}{\mathrm{~d}(\mathrm{r}-1)}\left(\mathrm{V} _{\mathrm{n}}-\mathrm{V} _{\mathrm{n}-1}\right) \Rightarrow \mathrm{S} _{\mathrm{n}}=\frac{-1}{\mathrm{~d}(\mathrm{r}-1)}\left(\mathrm{V} _{\mathrm{n}}-\mathrm{V} _{0}\right)$

E.g $\frac{1}{1.2}+\frac{1}{1.3}+\ldots \ldots \ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}$

$ \begin{aligned} & \frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\right) \\ & \frac{1}{1.2 .3 .4}+\frac{1}{2.3 .4 .5}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left(\frac{1}{1.2 .3}-\frac{1}{(n+1)(n+2)(n+3)}\right) \end{aligned} $

2. If $S _{n}=a _{1} a _{2} \ldots \ldots a _{r}+a _{2} a _{3} \ldots \ldots a _{r+1}+\ldots \ldots+a _{n} a _{n+1} \ldots \ldots a _{n+r-1}$

$ \begin{aligned} & T _{n}=a _{n} a _{n+1} \ldots \ldots . a _{n+r-1} \quad V _{n}=a _{n} a _{n+1} \ldots \ldots . a _{n+r-1} a _{n-r} \\ & V _{n}-V _{n-1}=a _{n} a _{n+1} \ldots . a _{n+r}-a _{n-1} a _{n} \ldots \ldots . a _{n+r-1} \\ & =\quad a _{n} a _{n+1} \ldots . a _{n+r-1}\left(a _{n+r}-a _{n-1}\right) \\ & =\quad T _{n}(r+1) d \\ & T _{n}=\frac{1}{d(r+1)}\left(V _{n}-V _{n-1}\right) \Rightarrow S _{n}=\frac{1}{d(r+1)}\left(V _{n}-V _{0}\right) \\ & 1.2+2.3+\ldots . .+n(n+1)=n(n+1) \frac{(n+2)}{3} \end{aligned} $

$1 \cdot 2 \cdot 3 \cdot 4+2 \cdot 3 \cdot 4 \cdot 5+\ldots \ldots \ldots+n(n+1)(n+2)(n+3)=n(n+1)(n+2)(n+3) \frac{(n+4)}{5}$

Alternative method

1. $\mathrm{S}=\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\ldots \ldots \ldots \ldots+\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}$

$ \begin{aligned} & =\quad \frac{1}{2}\left\{\frac{3-1}{1.2 .3}+\frac{4-2}{2.3 .4}+\ldots \ldots \ldots \ldots+\frac{(n+2)-n}{n(n+1)(n+2)}\right\} \\ & =\quad \frac{1}{2}\left\{\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\ldots \ldots+\left(\frac{1}{\mathrm{n}(\mathrm{n}+1)}-\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\right)\right\} \\ & =\frac{1}{2}\left\{\frac{1}{1.2}-\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\right\} \end{aligned} $

2. $1.2 .3+2.3 .4+\ldots \ldots+n(\mathrm{n}+1)(\mathrm{n}+2)=\frac{(\mathrm{n})(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{(3+1)}$

$=\frac{\mathrm{t} _{\mathrm{n}}(\mathrm{n}+3)}{3+1}$

i.e. Sum of the product of $m$ consecutive natural $S _{n}=\frac{T _{n}(n+m)}{m+1}$

Practice questions

1. If $R \subset(0, \pi)$ denote the set of values which satisfies the equation

$2^{1+|\cos x|+\left|\cos ^2 x\right|+\left|\cos ^3 x\right|+\ldots \ldots}=4$, then $R=$

(a). $\left\{\frac{-\pi}{3}\right\}$

(b). $\left\{\frac{\pi}{3}, \frac{2 \pi}{3}\right\}$

(c). $\left\{\frac{-\pi}{3}, \frac{2 \pi}{3}\right\}$

(d). $\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}\right\}$

2. Find the value of the expression $\sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{i}} \sum _{\mathrm{k}=1}^{\mathrm{j}} 1$

3. If in a series $\mathrm{t} _{\mathrm{n}}=\frac{\mathrm{n}}{(\mathrm{n}+1) !}$, Then $\sum _{\mathrm{n}=1}^{20} \mathrm{t} _{\mathrm{n}}$ is equal to

(a). $\frac{20 !-1}{20 !}$

(b). $\frac{21 !-1}{21 !}$

(c). $\frac{1}{2(n-1) !}$

(d). none of these

4. The value of $(0.2)^{\log _{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \ldots \infty\right)}$ is

(a). $1$

(b). $2$

(c). $\frac{1}{2}$

(d). $4$

5. $\lim _{n \rightarrow \infty}\left(1+3^{-1}\right)\left(1+3^{-2}\right)\left(1+3^{-4}\right)\left(1+3^{-8}\right) \ldots \ldots . .\left(1+3^{-2^{n}}\right)$ is equal to

(a). $1$

(b). $\frac{1}{2}$

(c). $\frac{3}{2}$

(d). none of these

6. $\lim _{\mathrm{n} \rightarrow \infty} \frac{1.2 .3+2.3 .4+3.4 .5+\ldots \ldots \ldots . . . \mathrm{upto} \mathrm{n} \text { terms }}{\mathrm{n}(1.2+2.3+3.4+\ldots \ldots \ldots \text {. }}$ is equal to $\mathrm{n}$ terms $)$

(a). $\frac{3}{4}$

(b). $\frac{1}{4}$

(c). $\frac{1}{2}$

(d). $\frac{5}{4}$

7. Sum to infinite terms of the series $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{9}+\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{2}{25}+\tan ^{-1} \frac{1}{18}+\ldots \ldots$.

(a). $\tan ^{-1} 3$

(b). $ \cot ^{-1} \frac{1}{3}$

(c). $\tan ^{-1} \frac{1}{3}$

(d). $\cot ^{-1} 3$

8. If $f(x)=a _{0}+a _{1} x+a _{2} x^{2}+\ldots \ldots+a _{n} x^{n}+\ldots \ldots$. and $\frac{f(x)}{1-x}=b _{0}+b _{1} x+b _{2} x^{2}+\ldots \ldots .+b _{n} x^{n}+\ldots \ldots$. If $a _{0}=1$ and $\mathrm{b} _{1}=3$ and $\mathrm{b} _{10}=\mathrm{k}^{11}-1$, and $\mathrm{a} _{0}, \mathrm{a} _{1}, \mathrm{a} _{2}, \ldots .$. are in G.P then $\mathrm{k}$ is

(a). $2$

(b). $3$

(c). $\frac{1}{2}$

(d). none of these

9. The value of $n$ for which $704+\frac{1}{2}(704)+\frac{1}{4}(704)+\ldots . .$. up to $n$ terms = $1984-\frac{1}{2}(1984)+\frac{1}{4}(1984) \ldots \ldots .$. upto $n$ terms is

(a). 5

(b). 3

(c). 4

(d). 10

10. If $1^{2}+2^{2}+3^{2}+\ldots \ldots+2003^{2}=(2003)(4007)(334)$ and (1) (2003) $+2(2002)+(3)(2001) \ldots \ldots .$. $(2003)(1)=(2003)(334)(x)$, then $x$ equals

(a). 2005

(b). 2004

(c). 2003

(d). 2001

11. Read the passage and answer the following questions

Let $\mathrm{T} _{1}, \mathrm{~T} _{2} \ldots . . \mathrm{T} _{\mathrm{n}}$ be the terms of a sequence and let $\left(\mathrm{T} _{2}-\mathrm{T} _{1}\right)=\mathrm{T} _{2}{ }^{1},\left(\mathrm{~T} _{3}-\mathrm{T} _{2}\right)=\mathrm{T} _{2}{ }^{1}, \ldots \ldots$. $\left(\mathrm{T} _{\mathrm{n}}-\mathrm{T} _{\mathrm{n}-1}\right)=\mathrm{T}^{1 \mathrm{n}-1}$

Case I: If $\mathrm{T} _{1}{ }^{1}, \mathrm{~T} _{2}{ }^{1} \ldots \ldots . \mathrm{T} _{\mathrm{n}-1}{ }^{1}$ are in A.P., then $\mathrm{T} _{\mathrm{n}}$ is quardratic in ’ $\mathrm{n}$ ‘. If $\mathrm{T} _{1}{ }^{1}-\mathrm{T} _{2}{ }^{1}$, $\mathrm{T} _{2}{ }^{1}-\mathrm{T} _{3}{ }^{1}$,…. are in A.P., then $\mathrm{T} _{\mathrm{n}}$ is cubic is $\mathrm{n}$.

Case II: If $T _{1}{ }^{1}, T _{2}{ }^{1} \ldots \ldots T _{n-1}^{1}$ are not in A.P., but in G.P. , then $T _{n}=a r^{n}+b$ where $r$ is the common ratio of the G.P. $\mathrm{T} _{1}{ }^{1}, \mathrm{~T} _{2}{ } _{2}{ }^{1} \mathrm{~T} _{3}{ }^{1} \ldots$. and $\mathrm{a}, \mathrm{b} \in \mathrm{R}$. Again if $\mathrm{T} _{1}{ }^{1}, \mathrm{~T} _{2}{ }^{1} \ldots \ldots \mathrm{T} _{\mathrm{n}-1}^{1}$ are not in G.P., but $\mathrm{T} _{2}{ }^{1}-\mathrm{T} _{1}{ } _{1}{ }^{1} \mathrm{~T} _{3}{ }^{1}-\mathrm{T} _{2}{ }^{1}, \ldots \ldots$ are in G.P., then $\mathrm{T} _{\mathrm{n}}$ is of the form $\mathrm{ar}^{\mathrm{n}-1}+\mathrm{bn}+\mathrm{c}$ and $\mathrm{r}$ is the C.R. of the G.P. $T _{2}{ }^{2}-T _{1}{ }^{1}, T _{3}{ }^{1}-T _{2}{ }^{1}, \ldots \ldots$ and a,b,c $\in$ R.

i. The sum of 20 terms of the series $3+7+14+24+37+\ldots \ldots$ is

(a). 4010

(b). 3860

(c). 4240

(d). none of these

ii. The $100^{\text {th }}$ term of the series $3+8+22+72+226+1036+\ldots$. . is divisible by $2^{\text {n }}$, then maxi mum value of $n$ is

(a). 4

(b). 2

(c). 3

(d). 5

iii. For the series $2+12+36+80+150+252+\ldots .$. , the value of $\lim _{n \rightarrow \infty} \frac{T _{n}}{n^{3}}$ is (where $T _{n}$ is the $n^{\text {th }}$ term)

(a). $2$

(b). $\frac{1}{2}$

(c). $1$

(d). none of these

12. Match the following