# Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progression Exercise 9.1

**Problem 1: Write the first terms of each of the following sequences whose nth term are:**

**(i) a**_{n} = 3n + 2

_{n}= 3n + 2

**Solution:**

Given:a

_{n}= 3n + 2By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{1 }= (3 × 1) + 2 = 3 + 2 = 5a

_{2}= (3 × 2) + 2 = 6 + 2 = 8a

_{3}= (3 × 3) + 2 = 9 + 2 = 11a

_{4}= (3 × 4) + 2 = 12 + 2 = 14a

_{5}= (3 × 5) + 2 = 15 + 2 = 17

∴ The first five terms are 5, 8, 11, 14, 17

**(ii) a**_{n} = (n – 2)/3

_{n}= (n – 2)/3

**Solution: **

Given:By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{n}= (n – 2)/3a

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

_{2}= (2 – 2)/3 = 0a

_{3}= (3 – 2)/3 = 1/3a

_{4}= (4 – 2)/3 = 2/3a

_{5}= (5 – 2)/3 = 3/3 =1

∴ The first five terms are -1/3, 0, 1/3, 2/3, 1

**(iii) a**_{n} = 3^{n}

_{n}= 3

^{n}

**Solution:**

Given:By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{1}= 3^{1}= 3;a

_{2}= 3^{2}= 9;a

_{3}= 3^{3 }= 27^{ };a

_{4}= 3^{4}= 81;a

_{5}= 3^{5}= 243.

∴ The first five terms are 3, 9, 27, 81, 243.

**(iv) a**_{n} = (3n – 2)/ 5

_{n}= (3n – 2)/ 5

**Solution:**

Given:By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{1}= (3 * 1 – 2)/5 = 1/5a

_{2}= (3 * 2 – 2)/5 = 4/5a

_{3}= (3 * 3 – 2)/5 = 7/5a

_{4}= (3* 4 – 2)/5= 10/5 =2a

_{5}= (3 * 5 – 2)/5 =13/5

∴ The first five terms are 1/5, 4/5, 7/5, 2, 13/5

**(v) a**_{n} = (-1)^{n}2^{n}

_{n}= (-1)

^{n}2

^{n}

**Solution:**

Given:By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{1}= (-1)^{1 }. 2^{1 }= -2a

_{2}= (-1)^{2 }. 2^{2 }= 4a

_{3}= (-1)^{3 }. 2^{3 }= -8a

_{4}= (-1)^{4 }. 2^{4 }= 16a

_{5}= (-1)^{5}. 2^{5 }= -32

∴ The first five terms are -2, 4, -8, 16, -32

**(vi) a**_{n} = n(n – 2)/2

_{n}= n(n – 2)/2

**Solution:**

Given:By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{1 }= (1.( 1 – 2))/2 = -1/2a

_{2}= (2.(2 – 2))/2 = 0a

_{3}= (3.(3 – 2))/2 = 3/2a

_{4}= (4.(4 – 2))/2 = 4a

_{5}= (5.(5 – 2))/2 =15/2

∴ The first five terms are -1/2, 0, 3/2, 4, 15/2

**(vii) a**_{n} = n^{2} – n + 1

_{n}= n

^{2}– n + 1

**Solution: **

Given:By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{1}= 1^{2}– 1 + 1 = 1a

_{2}= 2^{2 }– 2 + 1 = 3a

_{3}= 3^{2 }– 3 + 1 = 7a

_{4}= 4^{2 }– 4 + 1 = 13a

_{5}= 5^{2 }– 5 + 1 = 21

∴ The first five terms are 1, 3, 7, 13, 21

**(viii) a**_{n} = 2n^{2} – 3n + 1

_{n}= 2n

^{2}– 3n + 1

**Solution:**

Given:By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a

_{1}= 2 .1^{2 }– 3.1 + 1 = 0a

_{2}= 2. 2^{2 }– 3.2 + 1 = 3a

_{3}= 2.3^{2}– 3.3 + 1 = 10a

_{4}= 2.4^{2 }– 3.4 + 1 = 21a

_{5}= 2.5^{2 }– 3.5 + 1 = 36

∴ The first five terms are 0, 3, 10, 21, 36

**Problem 2: Find the indicated terms in each of the following sequences whose nth terms are:**

**(i) a**_{n} = 5n – 4; a_{12} and a_{15}

_{n}= 5n – 4; a

_{12}and a

_{15}

**Solution :**

Given:a

_{n}= 5n – 4a

_{12 }= By putting n=12a

_{12 }= 5.12 – 4 = 56a

_{15}= By putting n=15a

_{15}= 5.15 – 4 = 71

∴ The required terms a_{12}= 56, a_{15}= 71

**(ii) a**_{n} = (3n – 2)/(4n + 5), a_{7 }and a_{8}

_{n}= (3n – 2)/(4n + 5), a

_{7 }and a

_{8}

**Solution:**

Given:a

_{n}= (3n – 2)/(4n + 5)a

_{7}= By putting n=7= (3.7 – 2)/(4.7 + 5)

=19/33

a

_{8 }= By putting n = 8= (3.8 – 2)/(4.8 + 5)

= 2 2/37

∴ The required terms a_{7 }= 19/33, a_{8}= 22/37

** (iii) a**_{n} = n(n – 1)(n – 2); a_{5} and a_{8}

_{n}= n(n – 1)(n – 2); a

_{5}and a

_{8}

**Solution: **

Given:

a

_{n}= n(n – 1)(n – 2)By putting n=5

a

_{5}= 5(5 – 1).(5 – 2) = 5.4.3 = 60By putting n=8

a

_{8}= 8.(8 – 1).(8 – 2) = (8.7.6) = 336

∴ The required terms a_{5}= 60, a_{8}= 336

**(iv) a**_{n} = (n – 1)(2 – n)(3 + n); a_{1}, a_{2}, a_{3 }

_{n}= (n – 1)(2 – n)(3 + n); a

_{1}, a

_{2}, a

_{3 }

**Solution:**

Given:

a

_{n}= (n – 1)(2 – n)(3 + n)By putting n = 1

a

_{1}= (1 – 1)(2 – 1)(3 + 1) = 0.1.4 = 0By putting n = 2

a

_{2}= (2 – 1)(2 – 2)(3 + 2) = 1.0.5 = 0By putting n = 3

a

_{3}= (3 – 1)(2 – 3)(3 + 3) = 2.-1.6 = -12

∴ The required terms a_{1}= 0, a_{2}= 0 and a_{3 }= -12

**(v) a**_{n} = (-1)^{n}.n; a_{3}, a_{5}, a_{8}

_{n}= (-1)

^{n}.n; a

_{3}, a

_{5}, a

_{8}

**Solution:**

Given:

a

_{n}= (-1)^{n}.nBy putting n=3

a

_{3}= (-1)^{3}.3 = -1.3 = -3By putting n=5

a

_{5}= (-1)^{5}.5 = -1.5 = -5By putting n=8

a

_{8}= (-1)^{8}.8 = 1.8 = 8

∴ The required terms a_{3}= -3, a_{5}= -5 and a_{8}= 8

**Problem 3: Find the next five terms of each of the following sequence given by:**

**(i) a**_{1} = 1, a_{n} = a_{n – 1} + 2, n ≥ **2**

_{1}= 1, a

_{n}= a

_{n – 1}+ 2, n

**Solution:**

By putting n = 2, 3, 4, 5, 6 we get next five term of the sequence

a

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

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

_{4}= a_{3}+ 2 = 5 + 2 = 7a

_{5}= a_{4}+ 2 = 7 + 2 = 9a

_{6}= a_{5}+ 2 = 9 + 2 = 11

∴ The next five terms are 3, 5, 7, 9, 11

**(ii) a**_{1} = a_{2} = 2, a_{n} = a_{n – 1} – 3, n > 2

_{1}= a

_{2}= 2, a

_{n}= a

_{n – 1}– 3, n > 2

**Solution:**

By putting n = 3, 4, 5, 6, 7 we get next five term of the sequence

a

_{3}= a_{2}– 3 = 2 – 3 = -1a

_{4}= a_{3}– 3 = -1 – 3 = -4a

_{5}= a_{4}– 3 = -4 – 3 = -7a

_{6}= a_{5}– 3 = -7 – 3 = -10a

_{7}= a_{6}– 3 = -10 – 3 = -13

∴ The next five terms are -1, -4, -7, -10, -13

**(iii) a**_{1 }= -1, a_{n }= (a_{n – 1})/(n), n >= 2

_{1 }= -1, a

_{n }= (a

_{n – 1})/(n), n >= 2

**Solution:**

By putting n = 2, 3, 4, 5, 6 we get next five term of the sequence

a

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

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

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

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

_{6}= a_{5}/6 =(-1/120)/6 = -1/720

∴ The next five terms are -1/2, -1/6, -1/24, -1/120, -1/720

**(iv) a**_{1} = 4, a_{n} = 4a_{n – 1} + 3, n > 1

_{1}= 4, a

_{n}= 4a

_{n – 1}+ 3, n > 1

**Solution:**

By putting n = 2, 3, 4, 5, 6 we get next five term of the sequence

a

_{2 }= 4.a_{1}+ 3 = 4.4 + 3 = 19a

_{3}= 4.a_{2 }+ 3 = 4.19 + 3 = 79a

_{4}= 4.a_{3 }+ 3 = 4.79 + 3 = 319a

_{5}= 4.a_{4 }+ 3 = 4.319 + 3 = 1279a

_{6}= 4.a_{5 }+ 3 = 5.1279 + 3 = 5119

∴ The next five terms are 19, 79, 319, 1279, 5119