# Question 1 and 2 Exercise 4.1

Solutions of Question 1 and 2 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 1(i)

Classify into finite and infinite sequences

$2,4,6,8, \ldots ,50$

### Solution

It is finite sequence whose last term is $50 $.

## Question 1(ii)

Classify into finite and infinite sequences. $1,0,1,0,1, \ldots$.

### Solution

It is infinite sequence, the last term may be $0$ or $1$ , but how much terms in this sequence, we don't know.

## Question 1(iii)

Classify into finite and infinite sequences: $...,-4,0,4,8, \ldots, 60$

### Solution

This is infinite sequence.

## Question 1(iv)

Classify into finite and infinite sequences. $1,-\dfrac{1}{3}, \dfrac{1}{9},-\dfrac{1}{27}, \ldots,-\dfrac{1}{2187}$

### Solution

This finite sequence.

## Question 2(i)

Find first four terms of the sequence with the given general terms: $a_n=\dfrac{n(n+1)}{2}$

### Solution

Given: $$a_n=\dfrac{n(n+1)}{2}$$ For first term, put $n=1,$ $$a_1=\dfrac{1(1+1)}{2}=1$$ For second term, put $n=2$. $$a_2=\dfrac{2(2+1)}{2}=3$$ For third term, put $n=3,$ $$a_3=\dfrac{3(3+1)}{2}=6$$ For forth term, put $n=4,$ $$a_4=\dfrac{4(4+1)}{2}=10$$ Hence first four terms of the sequence are $1,3,6, 10$.

## Question 2(ii)

Find first four terms of the sequence with the given general terms: $a_n=(-1)^{n-1} 2^{n+1}$

### Solution

Given: $$a_n=(-1)^{n-1} 2^{n+1}$$ For first term, put $n=1$, then $$a_1=(-1)^{1-1} 2^{1+1}=2^2=4$$ For second term, put $n=2$, then $$a_2=(-1)^{2-1} 2^{2+1}=-2^3=-8$$ For third term, put $n=3$, then $$a_3=(-1)^{3-1} 2^{3-1}=2^4=16$$ For fourth term, put $n=4$, then $$a_4=(-1)^{4 \cdots 1} 2^{4+1}=-2^5=-32 \text {. }$$ Hence first four terms of the sequence are $4,-8,16,-32$.

## Question 2(iii)

Find first four terms of the sequence with the given general terms: $a_n =(\dfrac{1}{3})^n$

### Solution

Given: $$a_n =(\dfrac{1}{3})^n$$ For first term, put $n=1$, $$a_1=(\dfrac{1}{3})^1=\dfrac{1}{3}$$ For second term, put $n=2$, $$a_2=(\dfrac{1}{3})^2=\dfrac{1}{9}$$ For third term, put $n=3$, $$a_3=(\dfrac{1}{3})^3=\dfrac{1}{27}$$ For fourth term, put $n=4$, $$a_4=(\dfrac{1}{3})^4=\dfrac{1}{81}$$ Hence first four terms of the sequence are $\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{27},\dfrac{1}{81}$.

## Question 2(iv)

Find first four terms of the sequence with the given general terms: $a_n=\dfrac{n(n-1)(n-2)}{6}$

### Solution

Given: $$a_n=\dfrac{n(n-1)(n-2)}{6}$$ For first term, put $n=1,$ $$a_1=0$$ For second term, put $n=2$, $$a_2=0$$ For third term, put $n=3$, $$a_3=1$$ For fourth term, put $n=4$, $$a_4=4$$ Hence first four terms of the sequence are $0,0,1,4$.

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