### Question 1. Write the first five terms of a sequence whose n^{th }term is a_{n}=n(n+2).

**Solution: **

Given,

a_{n }= n(n + 2)Putting value of n as 1, 2, 3, 4 and 5. We get,

a

_{1}=1(1+2)=1(3)=3a

_{2}=2(2+2)=2(4)=8a

_{3}=3(3+2)=3(5)=15a

_{4}=4(4+2)=4(6)=24a

_{5}=5(5+2)=5(7)=35

Therefore, the first 5 terms of given series are 3, 8, 15, 24 and 35.

### Question 2. Write the first 5 terms of the series whose n^{th} term is a_{n} = n/(n + 1) .

**Solution:**

Given,

a_{n}= n/(n+1)Putting values of n as 1,2,3,4 and 5. We get,

a

_{1}=1/(1+1)=1/2a

_{2}=2/(2+1)=2/3a

_{3}=3/(3+1)=3/4a

_{4}=4/(4+1)=4/5a

_{5}=5/(5+1)=5/6

Therefore, the first 5 terms of the given series are 1/2 , 2/3 , 3/4 , 4/5 and 5/6.

### Question 3. Write the first five terms of series whose n^{th} term is a_{n }= 2^{n}.

**Solution:**

Given,

a_{n}=2^{n}Putting value of n as 1, 2, 3, 4 and 5. We get,

a

_{1}=2^{1}=2a

_{2}=2^{2}=4a

_{3}=2^{3}=8a

_{4}=2^{4}=16a

_{5}=2^{5}=32

Therefore, the first 5 terms of the given series are 2, 4, 8, 16 and 32.

### Question 4. Write the first five terms of series whose n^{th} term is a_{n }= (2n – 3)/6

**Solution:**

Given,

a_{n}=(2n -3)/6Putting value of n as 1, 2, 3, 4 and 5. We get,

a

_{1}=(2(1) -3)/6=-1/6a

_{2}=(2(2) -3)/6=1/6a

_{3}=(2(3) -3)/6=1/2a

_{4}=(2(4) -3)/6=5/6a

_{5}=(2(5) -3)/6=7/6

Therefore, the first 5 terms of the given series are -1/6, 1/6, 1/2, 5/6 and 7/6.

### Question 5. Write the first 5 terms of the sequence whose n^{th }term is a_{n} = (-1)^{n-1}5^{n+1}.

**Solution:**

Given,

a_{n}= (-1)^{n-1}5^{n+1}Putting values of n as 1, 2, 3, 4 and 5. We get,

a_{1 }= (-1)^{1-1}5^{1+1}=(-1)^{0}5^{2}=25a

_{2 }= (-1)^{2-1}5^{2+1}=(-1)^{1}5^{3}=-125a

_{3 }= (-1)^{3-1}5^{3+1}=(-1)^{2}5^{4}=625a

_{4 }= (-1)^{4-1}5^{4+1}=(-1)^{3}5^{5}=-3125a

_{5 }= (-1)^{5-1}5^{5+1}=(-1)^{4}5^{6}=15625

Therefore, the first 5 terms of the series are 25, -125, 625, -3125 and 15625.

### Question 6. Find the first five terms of the series whose n^{th} term is given as a_{n }= n(n^{2}+5)/4.

**Solution:**

Given, a

_{n }= n(n^{2}+5)/4Putting values of n as 1, 2, 3, 4 and 5. We get,

a

_{1 }= 1(1^{2}+5)/4=1(1+5)/4=1(6)/4=3/2a

_{2 }= 2(2^{2}+5)/4=2(4+5)/4=2(9)/4=9/2a

_{3 }= 3(3^{2}+5)/4=3(9+5)/4=3(14)/4=3(7)/2=21/2a

_{4 }= 4(4^{2}+5)/4=4(16+5)/4=4(21)/4=21a

_{5 }= 5(5^{2}+5)/4=5(25+5)/4=5(30)/4=5(15)/2=75/2

Therefore, the first 5 terms of the series are 3/2, 9/2, 21/2, 21 and 75/2.

### Question 7. Find the 17^{th} term of the sequence whose n^{th} term is given as a_{n}=4n – 3.

**Solution:**

Given,

a_{n }= 4n – 3Putting value of n as 17. We get,

a

_{17 }= 4(17) -3 = 68 – 3=65

Therefore, the 17^{th}term is 65.

### Question 8. Find the 7^{th} term of the sequence whose n^{th} term is given as a_{n }= n^{2}/(2n).

**Solution:**

Substituting n = 7. We get,

a

_{7 }= 7^{2}/(2*7)=49/14=7/2.

Therefore, the 7^{th}term is7/2.

### Question 9. Find the 9^{th} term of the sequence whose n^{th} term is given as a_{n }= (-1)^{n-1}n^{3}.

**Solution:**

Substituting n = 9. We get,

a

_{9 }= (-1)^{9-1}9^{3}=(-1)^{8}9^{3}=729.

Therefore, the 9^{th}term is 729.

### Question 10. Find the 20^{th} term of the sequence whose n^{th} term is gives as a_{n }= (n(n-2))/(n+3).

**Solution:**

Given,

a_{n }= (n(n-2))/(n+3)Putting value of n as 20. We get,

a

_{20 }= (20(20 – 2))/(20+3)=(20(18))/(23)=360/23

Therefore, the 20^{th}term is 360/23.

### Question 11. Find the first 5 terms of the following sequence. a_{1 }= 3, a_{n }= 3a_{n-1}+ 2 for all n>1.

**Solution:**

a

_{1 }= 3a

_{2 }= 3a_{2-1}+2=3a_{1}+2=3*3+2=9+2=11a

_{3 }= 3a_{3-1}+2=3a_{2}+2=3*11+2=33+2=35a

_{4 }= 3a_{4-1}+2=3a_{3}+2=3*35+2=105+2=107a

_{5 }= 3a_{5-1}+2=3a_{4}+2=3*107+2=321+2=323

The first 5 terms are 3, 11, 35, 107 and 323.

### Question 12. Write the first 5 terms of the following sequence. a_{1 }= -1, a_{n }= a_{n-1}/n for all n>1.

**Solution:**

a

_{1}=-1a

_{2}=a_{2-1}/2=a_{1}/2=-1/2a

_{3}=a_{3-1}/3=a_{2}/3=(-1/2)/3=-1/6a

_{4}=a_{4-1}/4=a_{3}/4=(-1/6)/4=-1/24a

_{5}=a_{5-1}/5=a_{4}/5=(-1/24)/5=-1/120

Therefore, the first 5 terms of the series are -1,-1/2, -1/6, -1/24 and -1/120.

### Question 13. Write the first 5 terms of the following sequence . a_{1 }= a_{2 }= 2, a_{n }= a_{n-1}-1 for all n > 2.

**Solution:**

a

_{1 }= 2a

_{2 }= 2a

_{3 }= a_{3-1}-1=a_{2}-1=2-1=1a

_{4 }= a_{4-1}-1=a_{3}-1=1-1=0a

_{5 }= a_{5-1}-1=a_{4}-1=0-1=-1

Therefore, the first 5 terms of the series are 2, 2, 1, 0 and -1.

### Question 14. The Fibonacci sequence is given by, 1 = a_{1 }= a_{2} and a_{n }= a_{n-1 }+ a_{n-2}, n > 2. Find a_{n+1}/a_{n} for the first 5 terms.

**Solution:**

a

_{1 }= 1, a_{2 }= 1a

_{3 }= a_{1}+a_{2}=1+1=2a

_{4 }= a_{2}+a_{3}=1+2=3a

_{5 }= a_{3}+a_{4}=2+3=5a

_{6 }= a_{4}+a_{5}=3+5=8Value of a

_{n+1}/a_{n}for,n = 1 is a

_{1+1}/a_{1}=a_{2}/a_{1}=1/1=1n = 2 is a

_{2+1}/a_{2}=a_{3}/a_{2}=2/1=2n = 3 is a

_{3+1}/a_{3}=a_{4}/a_{3}=3/2n = 4 is a

_{4+1}/a_{4}=a_{5}/a_{4}=5/3n = 5 is a

_{5+1}/a_{5}=a_{6}/a_{5}=8/5

Therefore, the required answer is 1, 2, 3/2, 5/3 and 8/5