## Chapter 3 – Arithmetic Progression

**Practice set 3.1**

**1. Which of the following sequences are A.P. ? If they are A.P. find the common difference. **

**(1) 2, 4, 6, 8, …**

**Solution:**

Here,

t₁ = 2, t₂ = 4, t₃ = 6, t₄ = 8

t₂ – t₁ = 4 – 2 = 2

t₃ – t₂ = 6 – 4 = 2

t₄ – t₃ = 8 – 6 = 2

This shows that the difference between any two consecutive terms is constant.

Hence, the given sequence is an A.P. and common difference (d) is 2.

**(2) 2, \(\large \frac {5}{2}\), 3, \(\large \frac {7}{3}\), …**

**Solution:**

Here,

t₁ = 2, t₂ = \(\large \frac {5}{2}\), t₃ = 3, t₄ = \(\large \frac {7}{3}\)

t₂ – t₁ = \(\large \frac {5}{2}\) – 2 = \(\large \frac {5\,–\,4}{2}\) = \(\large \frac {1}{2}\)

t₃ – t₂ = 3 – \(\large \frac {5}{2}\) = \(\large \frac {6\,–\,5}{2}\) = \(\large \frac {1}{2}\)

t₄ – t₃ = \(\large \frac {7}{3}\) – 3 = \(\large \frac {7\,–\,9}{3}\) = \(\large \frac {–\,2}{3}\)

This shows that the difference between any two consecutive terms is not constant.

Hence, the given sequence is not an A.P.

**(3) – 10, – 6, – 2, 2, …**

**Solution:**

Here,

t₁ = – 10, t₂ = – 6, t₃ = – 2, t₄ = 2

t₂ – t₁ = – 6 – (– 10) = – 6 + 10 = 4

t₃ – t₂ = – 2 – (– 6) = – 2 + 6 = 4

t₄ – t₃ = 2 – (– 2) = 2 + 2 = 4

This shows that the difference between any two consecutive terms is constant.

Hence, the given sequence is an A.P. and the common difference (d) is 4.

**(4) 0.3, 0.33, .0333, …**

**Solution:**

Here,

t₁ = 0.3, t₂ = 0.33 , t₃ = 0.333

t₂ – t₁ = 0.33 – 0.3 = 0.03

t₃ – t₂ = 0.333 – 0.33 = 0.003

This shows that the difference between any two consecutive terms is not constant.

Hence, the given sequence is not an A.P.

**(5) 0, – 4, – 8, – 12, …**

**Solution:**

Here,

t₁ = 0 , t₂ = – 4, t₃ = – 8, t₄ = –12

t₂ – t₁ = – 4 – 0 = – 4

t₃ – t₂ = – 8 – (– 4) = – 8 + 4 = – 4

t₄ – t₃ = –12 – (– 8) = – 12 + 8 = – 4

This shows that the difference between any two consecutive terms is constant.

Hence, the given sequence is an A.P. and the common difference (d) is – 4.

**(6) – \(\large \frac {1}{5}\), – \(\large \frac {1}{5}\), – \(\large \frac {1}{5}\), …**

**Solution:**

Here,

t₁ = – \(\large \frac {1}{5}\), t₂ = – \(\large \frac {1}{5}\), t₃ = – \(\large \frac {1}{5}\)

t₂ – t₁ = – \(\large \frac {1}{5}\) – \(\(\)– \(\large \frac {1}{5}\)\(\)\) = – \(\large \frac {1}{5}\) + \(\large \frac {1}{5}\) = 0

t₃ – t₂ = – \(\large \frac {1}{5}\) – \(\(\)– \(\large \frac {1}{5}\)\(\)\) = – \(\large \frac {1}{5}\) + \(\large \frac {1}{5}\) = 0

This shows that the difference between any two consecutive terms is constant.

Hence, the given sequence is an A.P. and the common difference (d) is zero.

**(7) 3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\), …**

**Solution:**

Here,

t₁ = 3, t₂ = 3 + \(\sqrt{2}\), t₃ = 3 + 2\(\sqrt{2}\), t₄ = 3 + 3\(\sqrt{2}\)

t₂ – t₁ = 3 + \(\sqrt{2}\) − 3 = \(\sqrt{2}\)

t₃ – t₂ = 3 + 2\(\sqrt{2}\) – (3 + 2\(\sqrt{2}\)) = 3 + 2\(\sqrt{2}\) – 3 – \(\sqrt{2}\) = 2\(\sqrt{2}\) – \(\sqrt{2}\) = \(\sqrt{2}\)

t₄ – t₃ = 3 + 3\(\sqrt{2}\) – (3 + 2\(\sqrt{2}\)) = 3 + 3\(\sqrt{2}\) – 3 – 2\(\sqrt{2}\) = 3\(\sqrt{2}\) – 2\(\sqrt{2}\) = \(\sqrt{2}\)

This shows that the difference between any two consecutive terms is constant.

Hence, the given sequence is an A.P. and the common difference (d) is 2.

**(8) 127, 132, 137, …**

**Solution:**

Here,

t₁ = 127, t₂ = 132, t₃ = 137

t₂ – t₁ = 132 – 127 = 5

t₃ – t₂ = 137 – 132 = 5

This shows that the difference between any two consecutive terms is constant.

Hence, the given sequence is an A.P. with a common difference (d) is 5.

**2. Write an A.P. whose first term is a and the common difference is d in each of the following.**

**(1) a = 10, d = 5 **

**Solution:**

a = 10, d = 5

t₁ = 10

t₂ = t₁ + d = 10 + 5 = 15

t₃ = t₂ + d = 15 + 5 = 20

t₄ = t₃ + d = 20 + 5 = 25

∴ Arithmetic progression is 10, 15, 20, 25, …

**(2) a = – 3, d = 0 **

**Solution:**

a = – 3, d = 0

t₁ = – 3

t₂ = t₁ + d = – 3 + 0 = – 3

t₃ = t₂ + d = – 3 + 0 = – 3

t₄ = t₃ + d = – 3 + 0 = – 3

∴ Arithmetic progression is –3, – 3, –3, – 3, …

**(3) a = – 7, d = \(\large \frac {1}{2}\)**

**Solution:**

a = – 7, d = \(\large \frac {1}{2}\)

t₁ = – 7

t₂ = t₁ + d = – 7 + \(\large \frac {1}{2}\) = – \(\large \frac {–\,14\,+\,1}{2}\) = – \(\large \frac {13}{2}\)

t₃ = t₂ + d = – \(\large \frac {13}{2}\) + \(\large \frac {1}{2}\) = – \(\large \frac {–\,13\,+\,1}{2}\) = – \(\large \frac {12}{2}\) = – 6

t₄ = t₃ + d = – 6 + \(\large \frac {1}{2}\) = – \(\large \frac {–\,12\,+\,1}{2}\) = – \(\large \frac {11}{2}\)

∴ Arithmetic progression is – 7, – \(\large \frac {13}{2}\), – 6, – \(\large \frac {11}{2}\), …

**(4) a = – 1.25, d = 3 **

**Solution:**

a = – 1.25 , d = 3

t₁ = – 1.25

t₂ = t₁ + d = – 1.25 + 3 = 1.75

t₃ = t₂ + d = 1.75 + 3 = 4.75

t₄ = t₃ + d = 4.75 + 3 = 7.75

∴ Arithmetic progression is – 1.25, 1.75, 4.75, 7.75, …

**(5) a = 6, d = – 3 **

**Solution:**

a = 6, d = – 3

t₁ = 6

t₂ = t₁ + d = 6 + (– 3) = 6 – 3 = 3

t₃ = t₂ + d = 3 + (– 3) = 3 – 3 = 0

t₄ = t₃ + d = 0 + (– 3) = – 3

∴ Arithmetic progression is 6, 3, 0, – 3, ….

**(6) a = – 19, d = – 4**

**Solution:**

a = t₁ = – 19, d = – 4

t₁ = – 19

t₂ = t₁ + d = – 19 + (– 4) = – 19 – 4 = – 23

t₃ = t₂ + d = – 23 + (– 4) = – 23 – 4 = – 27

t₄ = t₃ + d = – 27 + (– 4) = – 27 – 4 = – 31

∴ Arithmetic progression is – 19,– 23, – 27, – 31, …

**3. Find the first term and common difference for each of the A.P.**

**(1) 5, 1, – 3, – 7, …**

**Solution:**

Here,

t₁ = 5, t₂ = 1, t₃ = – 3, t₄ = –7

For an A.P.,

a = t₁ = 5

d = tₙ₊₁ – tₙ

∴ d = t₂ – t₁ = 1 – 5 = – 4

∴ d = t₃ – t₂ = – 3 – 1 = –4

∴ The first term (a) is 5 and the common difference (d) is – 4.

**(2) 0.6, 0.9, 1.2, 1.5, … **

**Solution:**

Here,

t₁ = 0.6, t₂ = 0.9, t₃ = 1.2, t₄ = 1.5

For an A.P.,

a = t₁ = 0.6

d = tₙ + t n – 1

∴ d = t₂ – t₁ = 0.9 – 0.6 = 0.3

∴ d = t₃ – t₂ = 1.2 – 0.9 = 0.3

∴ The first term (a) is 0.6 and the common difference (d) is 0.3.

**(3) 127, 135, 143, 151, …**

**Solution:**

Here,

t₁ = 127, t₂ = 135, t₃ = 143, t₄ = 151

For an A.P.,

a = t₁ = 127

d = tₙ₊₁ – tₙ

∴ d = t₂ – t₁ = 135 – 127 = 8

∴ d = t₃ – t₂ = 143 – 135 = 8

∴ The first term (a) is 127 and the common difference (d) is 8.

**(4) \(\large \frac {1}{4}\), \(\large \frac {3}{4}\), \(\large \frac {5}{4}\), \(\large \frac {7}{4}\), …**

**Solution:**

Here,

t₁ = a = \(\large \frac {1}{4}\), t₂ = \(\large \frac {3}{4}\), t₃ = \(\large \frac {5}{4}\), t₄ = \(\large \frac {7}{4}\)

For an A. P. d = tₙ₊₁ – tₙ

d = t₂ – t₁ = \(\large \frac {3}{4}\) – \(\large \frac {1}{4}\) = \(\large \frac {3\,–\,1}{4}\) = = \(\large \frac {2}{4}\) = \(\large \frac {1}{2}\)

d = t₃ – t₂ = \(\large \frac {5}{4}\) – \(\large \frac {3}{4}\) = \(\large \frac {5\,–\,3}{4}\) = = \(\large \frac {2}{4}\) = \(\large \frac {1}{2}\)

∴ The first term (a) is \(\large \frac {1}{4}\) and the common difference (d) is \(\large \frac {1}{2}\).

**Practice set 3.2**

**1. Write the correct number in the given boxes from the following A. P.**

**1. Write the correct number in the given boxes from the following A. P.**

**(i) 1, 8, 15, 22, … **

**Here a = __, t₁ = __, t₂ = __, t₃ = __, **

**t₂ – t₁ = __ – __ = __**

**t₃ – t₂= __ – __= __**

**∴ d = __**

**∴ d = __**

**Solution:**

**a = 1, t₁ = 1, t₂ = 8, t₃ = 15, **

**t₂ – t₁ = 8 – 1 = 7**

**t₃ – t₂= 15 – 8 = 7**

**∴ d = 7**

** **

**(ii) 3, 6, 9, 12, … **

**Here t₁ = __, t₂ = __, t₃ = __, t₄ = __,**

**t₂ – t₁ = __**

**t₃ – t₂ = __**

**∴ d = __**

**∴ d = __**

**Solution:**

**t₁ = 3, t₂ = 6, t₃ = 9, t₄ = 12,**

**t₂ – t₁ = 6 – 3 = 3**

**t₃ – t₂ = 9 – 6 = 3**

**∴ d = 3**

** **

**(iii) – 3, – 8, – 13, – 18, … **

**Here t₃ = __, t₂ = __, t₄ = __, t₁ = __,**

**t₂ – t₁ = __, **

**t₃ – t₂ = __**

**∴ a = __, d = __**

**∴ a = __, d = __**

**Solution:**

**t₃ = – 13, t₂ = – 8, t₄ = – 18, t₁ = – 3,**

**t₂ – t₁ = – 8 – (– 3) = – 8 + 3 = – 5 **

**t₃ – t₂ = – 13 – (– 8) = – 13 + 8 = – 5 **

**∴ a = – 3, d = – 5**

** **

**(iv) 70, 60, 50, 40, … **

**Here t₁ = __, t₂ = __, t₃ = __, … **

**∴ a = __, d = __**

**∴ a = __, d = __**

**Solution:**

**t₁ = 70, t₂ = 60, t₃ = 50, … **

**∴ a = 70, d = 10**

**2. Decide whether the following sequence is an A.P., if so find the 20th term of the progression.**

**2. Decide whether the following sequence is an A.P., if so find the 20th term of the progression.**

**– 12, – 5, 2, 9, 16, 23, 30, … **

**– 12, – 5, 2, 9, 16, 23, 30, …**

**Solution:**

**Here t₁ = – 12, t₂ = –5 , t₃ = 2, t₄ = 9, …**

**t₂ – t₁ = –5 – (– 12) = – 5 + 12 = 7**

**t₃ – t₂ = 2 – (– 5) = 2 + 5 = 7**

**t₄ – t₃ = 9 – 2 = 7**

**This shows that the difference between any two consecutive terms is constant.**

**∴ The given sequence is an A.P.**

**Here a = – 12, d = 7, n = 20, t₂₀ = ?**

**tₙ = a + (n – 1) d**

**∴ t₂₀ = – 12 + (20 – 1) 7 **

**∴ t₂₀ = – 12 + 19 × 7**

**∴ t₂₀ = – 12 + 133**

**∴ t₂₀ = 121**

**Ans:** 20th term of the A.P. is 121.

**3. Given Arithmetic Progression 12, 16, 20, 24, … Find the 24th term of this progression. **

**3. Given Arithmetic Progression 12, 16, 20, 24, … Find the 24th term of this progression.**

**Solution:**

**The A.P. 12, 16, 20, 24……**

**Here a = 12, d = 16 – 12 = 4, n = 24, t₂₄ = ?**

**tₙ = a + (n – 1) d**

**∴ t₂₄ = 12 + (24 – 1) 4 **

**∴ t₂₄ = 12 + 23 × 4**

**∴ t₂₄ = 12 + 92**

**∴ t₂₄ = 104**

**Ans:** 24th term of the A.P. is 104.

**4. Find the 19th term of the following A.P. **

**4. Find the 19th term of the following A.P.**

**7, 13, 19, 25, … **

**Solution:**

**The A.P. is 7, 13, 19, 25, …**

**Here a = 7, d = 13 – 7 = 6, n = 19, t₁₉ = ?**

**t = a + (n – 1) d**

**t₁₉ = 7 + (19 – 1) 6 **

**∴ t₁₉ = 7 + 18 × 6**

**∴ t₁₉ = 7 + 108**

**∴ t₁₉ = 115**

**Ans:** 19th term of the A.P. is 115

**5. Find the 27th term of the following A.P. **

**5. Find the 27th term of the following A.P.**

**9, 4, – 1, – 6, – 11, … **

**9, 4, – 1, – 6, – 11, …**

**Solution:**

**The A.P. is 9, 4, – 1, – 6, – 11, …**

**Here, a = 9, d = 4 – 9 = –5, n = 27, t₂₇ = ?**

**tₙ = a + (n – 1) d**

**∴ t₂₇ = 9 + (27 – 1) (– 5) **

**∴ t₂₇= 9 + 26 × (– 5)**

**∴ t₂₇= 9 – 130**

**∴ t₂₇ = – 121**

**Ans:** 27th term of the A.P. is – 121

**6. Find how many three digit natural numbers are divisible by 5. **

**6. Find how many three digit natural numbers are divisible by 5.**

**Solution:**

**The three digit natural numbers divisible by 5 are 100, 105, 110, … , 995**

**Here, t₁ = 100, t₂ = 105, t₃ =110, …**

**t₂ – t₁ = 105 – 100 = 5**

**t₃ – t₂ = 110 – 105 = 5**

**Since the difference between any two consecutive terms is constant.**

**∴ The given sequence is an A.P., where a = 100, d = 5 **

**let tₙ = 995, n = ? **

**tₙ = a + (n – 1) d**

**∴ 995 = 100 + (n – 1) 5**

**∴ 895 = 5n – 5**

**∴ 895 + 5 = 5n**

**∴ 5n = 900**

**∴ n = \(\large \frac {900}{5}\)**

**∴ n = 180**

**Ans:** There are 180 three digit natural numbers which are divisible by 5.

**7. The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th term of that A.P. **

**7. The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th term of that A.P.**

**Solution:**

**For an A.P. **

**t₁₁ = 16 and t₂₁ = 29**

**t₄₁ = ?**

**tₙ = a + (n – 1) d**

**∴ t₁₁ = a + (11 – 1) d**

**∴ 16 = a + 10d**

**∴ a + 10d = 16 …(i)**

**Also,**

**t₂₁= a + (21 – 1) d**

**∴ 29 = a + 20d**

**∴ a + 20d = 29 …(ii)**

**Subtracting (i) from (ii),**

** a + 20d = 29**

** a + 10d = 16**

**(–) (–) (–) **

** ______________**

** 10d = 13**

** ∴ d = \(\large \frac {13}{10}\)**

**Substituting d = \(\large \frac {13}{10}\) in (i)**

**a + 10d = 16**

**∴ a + 10 × \(\large \frac {13}{10}\) = 16**

**∴ a + 13 = 16**

**∴ a = 16 – 13**

**∴ a = 3**

**Here, a = 3 ; d = \(\large \frac {13}{10}\), n = 41, t₄₁ = ? **

**tₙ = a + (n – 1) d**

**∴ t₄₁ = 3 + (41 – 1) \(\large \frac {13}{10}\)**

**∴ t₄₁ = 3 + 40 × \(\large \frac {13}{10}\)**

**∴ t₄₁ = 3 + 4 × 13**

**∴ t₄₁ = 3 + 52**

**∴ t₄₁ = 55**

**Ans:** 41st term of the A.P. is 55.

**8. 11, 8, 5, 2, … In this A.P. which term is number – 151? **

**8. 11, 8, 5, 2, … In this A.P. which term is number – 151?**

**Solution:**

**The A.P. is 11, 8, 5, 2, …**

**Here, a = 11, **

**d = t₂ – t₁ = 8 – 11 = – 3, **

**tₙ = – 151, n = ?**

**tₙ = a + (n – 1) d**

**∴ – 151 = 11 + (n – 1) (– 3)**

**∴ – 151 = 11 – 3n + 3**

**∴ – 151 = 14 – 3n**

**∴ 3n = 14 + 151**

**∴ 3n = 165**

**∴ n = \(\large \frac {165}{3}\)**

**∴ n = 55**

**Ans:** 55th term of the A.P. is – 151.

**9. In the natural numbers from 10 to 250, how many are divisible by 4? **

**9. In the natural numbers from 10 to 250, how many are divisible by 4?**

**Solution:**

**The natural numbers from 10 to 250 divisible by 4 are: **

**12, 16, 20, …. 248**

**Here t₁ = 12, t₂ = 16, t₃ = 20, …**

**t₂ – t₁ = 16 – 12 = 4 **

**t₃ – t₂ = 20 – 16 = 4**

**This shows that the difference between any two consecutive terms is constant.**

**∴ The given sequence is an A.P., where a = 12, d = 4, **

**let tₙ = 248, n = ?**

**tₙ = a + (n – 1) d**

**∴ 248 = 12 + (n – 1) 4**

**∴ 248 – 12 = 4n – 4**

**∴ 236 = 4n – 4**

**∴ 236 + 4 = 4n**

**∴ 4n = 240**

**∴ n = \(\large \frac {240}{4}\)**

**∴ n = 60**

**Ans:** There are 60 natural numbers from 10 to 250 which are divisible by 4.

**10. In an A.P. 17th term is 7 more than its 10th term. Find the common difference.**

**10. In an A.P. 17th term is 7 more than its 10th term. Find the common difference.**

**Solution:**

**tₙ = a + (n – 1) d**

**∴ t₁₇ = a + (17 – 1) d**

**∴ t₁₇ = a + 16d …(i)**

**Also, **

**t₁₀ = a + (10 – 1) d**

**t₁₀ = a + 9d …(ii)**

**Now, **

**t₁₇ = t₁₀ + 7 … [Given]**

**∴ a + 16d = a + 9d + 7 … [From (i) and (ii)]**

**∴ 7d = 7**

**∴ d = \(\large \frac {7}{7}\)**

**∴ d = 1 **

**Ans:** The common difference is 1.

**Practice set 3.3**

**Practice set 3.3**

**1. First term and common difference of an A.P. are 6 and 3 respectively ; find S₂₇.**

**1. First term and common difference of an A.P. are 6 and 3 respectively ; find S₂₇.****a = 6, d = 3, S₂₇ = ?**

**Sₙ = \(\large \frac {n}{2}\) [ __ + (n – 1) d]**

**S₂₇ = \(\large \frac {27}{2}\) [12 + (27 – 1) __]**

**S₂₇ = \(\large \frac {27}{2}\) × __ **

**S₂₇ = 27 × 45 **

**S₂₇ = __**

**Solution:**

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**S₂₇ = \(\large \frac {27}{2}\) [12 + (27 – 1) 3]**

**S₂₇ = \(\large \frac {27}{2}\) × 90 **

**S₂₇ = 27 × 45 **

**S₂₇ = 1215**

**2. Find the sum of the first 123 even natural numbers.**

**2. Find the sum of the first 123 even natural numbers.****Solution:**

**The first n even natural numbers are: **

**2, 4, 6, 8,…..**

**They form an A.P. with **

**a = 2, d = t₂ – t₁ = 4 – 2 = 2**

**We know, **

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₁₂₃ = \(\large \frac {123}{2}\) [2(2) + (123 – 1)2]**

**∴ S₁₂₃ = \(\large \frac {123}{2}\) [4 + 122(2)]**

**∴ S₁₂₃ = \(\large \frac {123}{2}\) [4 + 244]**

**∴ S₁₂₃ = \(\large \frac {123}{2}\) [248]**

**∴ S₁₂₃ = 123 [124]**

**∴ S₁₂₃ = 15252**

**Ans:** The sum of the first 123 even natural numbers in 15252.

**3. Find the sum of all even numbers between 1 and 350.**

**3. Find the sum of all even numbers between 1 and 350.****Solution:**

**The even natural numbers from 1 to 350 are 2, 4, 6, 8, … , 348.**

**They form an A.P. where a = 2, d = 4 – 2 = 2**

**Let tₙ = 348**

**Now, **

**tₙ = a + (n – 1) d**

**∴ 348 = 2 + (n – 1) 2**

**∴ 348 – 2 = 2n – 2**

**∴ 346 = 2n – 2**

**∴ 346 + 2 = 2n**

**∴ 2n = 348**

**∴ n = \(\large \frac {348}{2}\)**

**∴ n = 174**

**Now **

**Sₙ = \(\large \frac {n}{2}\) [t₁ + tₙ]**

**∴ Sₙ = \(\large \frac {174}{2}\) [2 + 348]**

**∴ Sₙ = \(\large \frac {174}{2}\) × 350**

**∴ Sₙ = 30450**

**Ans:** The sum of even natural numbers between 1 to 350 is 30450.

**4. In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.**

**4. In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.****Solution:**

**t₁₉ = 52 , and t₃₈ = 128 **

**S₅₆ = ?**

**tₙ = a + (n – 1) d**

**∴ t₁₉ = a + (19 – 1) d**

**∴ 52 = a + 18d**

**∴ a + 18d = 52 …(i)**

**Also, **

**t₃₈ = a + (38 – 1) d**

**∴ 128 = a + 37d**

**∴ a + 37d = 128 … (ii)**

**Adding equations (i) and (ii), we get,**

** a + 18d = 52**

**+ a + 37d = 128**

** ___________________**

**2a + 55d = 180 …(iii)**

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₅₆ = \(\large \frac {56}{2}\) [2a + (56 – 1) d]**

**∴ S₅₆ = 28 [2a + 55d]**

**∴ S₅₆ = 28 × 180 … [From (iii)]**

**∴ S₅₆ = 5040**

**Ans:** The required sum is 5040.

**5. Complete the following activity to find the sum of natural numbers between 1 and 140 which are divisible by 4.**

**5. Complete the following activity to find the sum of natural numbers between 1 and 140 which are divisible by 4.**

**Solution:**

**Working:**

**a = 4, d = 4, tₙ = 136**

**∴ tₙ = a + (n – 1) d**

**∴ 136 = 4 + (n – 1) 4**

**∴ 136 = 4 + 4n – 4**

**∴ n = \(\large \frac {136}{4}\)**

**∴ n = 34**

**S₃₄ = \(\large \frac {34}{2}\) [2 × 4 + (34 – 1) 4]**

**∴ S₃₄ = 17(8 + 33 × 4)**

**∴ S₃₄ = 17(140)**

**∴ S₃₄ = 2380**

**Ans:** The sum of numbers between 1 to 140, which are divisible by 4 is 2380.

**6. Sum of first 55 terms in an A.P. is 3300, find its 28th term.**

**6. Sum of first 55 terms in an A.P. is 3300, find its 28th term.****Solution:**

**Here, S₅₅ = 3300, t₂₈ = ?**

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d**

**∴ S₅₅ = \(\large \frac {55}{2}\) [2a + (55 – 1) d]**

**∴ 3300 = \(\large \frac {55}{2}\) [2a + 54d]**

**∴ 3300 = \(\large \frac {55}{2}\) × 2 [a + 27d]**

**∴ 3300 = 55 (a + 27d)**

**∴ a + 27d = \(\large \frac {3300}{55}\)**

**∴ a + 27d = 60 …(i)**

**Now, **

**tₙ = a + (n – 1) d**

**∴ t₂₈ = a + (28 – 1) d**

**∴ t₂₈ = a + 27d**

**∴ t₂₈ = 60 … [From (i)]**

**Ans:** The 28th term is 60.

**7. In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.**

**7. In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.****(Assume that three consecutive terms in A.P. are a – d , a, a + d)**

**(Assume that three consecutive terms in A.P. are a – d , a, a + d)****Solution:**

**Let the three consecutive terms be a – d, a, a + d.**

**Their sum is 27**

**∴ a – d + a + a + d = 27**

**∴ 3a = 27**

**∴ a = 9**

**The product of all three terms is 504.**

**∴ (a – d) × a × (a + d) = 504**

**∴ (a² – d²) × a = 504 … [Using (a + b) (a – b) = a² – b²]**

**∴ (9² – d²) × 9 = 504**

**∴ 81 – d² = \(\large \frac {504}{9}\)**

**∴ 81 – d² = 56**

**∴ d² = 81 – 56**

**∴ d² = 25**

**∴ d = ± 5 … [Taking square root]**

**When d = 5, a = 9, the required terms are a – d, a , a + d , i.e. 4, 9, 14**

**When d = – 5, a = 9, the required terms are a – d, a, a + d, i.e. 14, 9, 4**

**Ans:** The required terms are 4, 9, 14 or 14, 9, 4.

**8. Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14. **

**8. Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.****(Assume the four consecutive terms in A.P. are a – d , a, a + d, a + 2d)**

**(Assume the four consecutive terms in A.P. are a – d , a, a + d, a + 2d)****Solution:**

**Let four consecutive terms in A.P. are a – d, a, a + d, a + 2d**

**Their sum is 12**

**∴ a – d + a + a + d + a + 2d = 12**

**∴ 4a + 2d = 12**

**∴ 2(2a + d) = 12**

**∴ 2a + d = 6 …(i)**

**The sum of 3rd and 4th terms is 14.**

**∴ a + d + a + 2d = 14**

**∴ 2a + 3d = 14 …(ii)**

**Subtracting (ii) from (i),**

** 2a + d = 6**

** 2a + 3d = 14**

** (–) (–) (–) **

** _________________**

** – 2d = – 8**

**∴ d = \(\large \frac {–\,8}{–\,2}\)**

**∴ d = 4**

**Substituting d = 4 in (i) we get,**

**2a + d = 6**

**∴ 2a + 4 = 6**

**∴ 2a = 6 – 4 **

**∴ 2a = 2**

**∴ a = 1**

**Ans:** The required terms are a – d, a, a + d, a + 2d i.e – 3, 1, 5, 9.

**9. If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.**

**9. If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.****Given:**

**For an A.P., **

**t₉ = 0 **

**To prove: **

**t₂₉ = 2t₁₉**

**∴ t₂₉ – 2t₁₉ = 0**

**Proof:**

**tₙ = a + (n – 1) d**

**∴ t₉ = a + (9 – 1) d**

**∴ 0 = a + 8d**

**∴ a + 8d = 0 …(i)**

**t₂₉ = a + (29 – 1) d**

**∴ t₂₉ = a + 28d**

**∴ t₁₉ = a + (19 – 1) d**

**∴ t₁₉ = a + 18d**

**∴ t₂₉ – 2t₁₉ **

**= a + 28d – 2(a + 18d)**

**= a + 28d – 2a – 36d**

**= – a – 8d**

**= – (a + 8d)**

**= 0 … [From (i)]**

** **

**∴ t₂₉ – 2t₁₉ = 0**

**∴ t₂₉ = 2t₁₉**

**Hence proved.**

**Practice set 3.4**

**Practice set 3.4****1. On 1st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?**

**1. On 1st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?****Solution:**

**2016 is a leap year.**

**∴ Number of days from 1st January 2016 to 31st December 2016 = 366**

**The savings of Sanika per day, beginning with the first day, are 10, 11, 12, … for 366 days. These every day savings form an A.P. with first days saving (a) = 10**

**Difference in saving made in two successive days, (d) = 1**

**Total number of days in the year 2016 (n) = 366**

**∴ Total saving for the year 2016 (S₃₆₆) = ?**

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₃₆₆ = [2 × 10 + (366 – 1)1]**

**∴ S₃₆₆ = 183 [20 + 365]**

**∴ S₃₆₆ = 183 × 385**

**∴ S₃₆₆ = 70,455**

**Ans:** Sanika saved ₹ 70,455 in the year 2016.

**2. A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly installments. Each installment being less than the preceding one by ₹ 40. Find the amount of the first and last installment.**

**2. A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly installments. Each installment being less than the preceding one by ₹ 40. Find the amount of the first and last installment.****Solution:**

**Each installment is less than the preceding investment by ₹ 40.**

**Therefore the installments are in A.P.**

**Total money repaid in 12 installments (S₁₂) = 8000 + 1360 = ₹ 9360**

**No. of installments (n) = 12**

**Difference between two consecutive installments (d) = – 40**

** **

**First installment (a) = ?, Last installment (t₁₂) = ?**

** **

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₁₂ = \(\large \frac {12}{2}\) [2a + (12 – 1)(– 40)]**

**∴ 9360 = 6 [2a + 11 × (– 40)]**

**∴ \(\large \frac {9360}{6}\) = 2a – 440**

**∴ 1560 = 2a – 440**

**∴ 1560 + 440 = 2a**

**∴ 2a = 2000**

**∴ a = \(\large \frac {2000}{2}\)**

**∴ a = 1000**

** **

**tₙ = a + (n – 1) d**

**∴ t₁₂ = 1000 + (12 – 1) (– 40)**

**∴ t₁₂ = 1000 + 11 × (– 40)**

**∴ t₁₂ = 1000 – 440**

**∴ t₁₂ = 560**

** **

**Ans:** First installment is ₹ 1000 and last installment is ₹ 560.

**3. Sachin invested in a national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.**

**3. Sachin invested in a national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.****Solution:**

**Yearly investments of Sachin are as follows:**

**5000, 7000, 9000, ….**

**The yearly investments form an A.P. with first year investment (a) = 5000**

**Difference between investments done in two successive years (d) = 2000**

**No. of years (n) = 12**

**Total investment done in 12 years S₁₂ = ?**

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₁₂ = \(\large \frac {12}{2}\) [2 × 5000 + (12 – 1) 2000]**

**∴ S₁₂ = 6 [10000 + 11 × 2000]**

**∴ S₁₂ = 6 × 32,000**

**∴ S₁₂ = 1,92,000**

** **

**Ans:** Total investment done in 12 years is ₹ 1,92,000

**4. There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?**

**4. There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?****Solution:**

**The number of seats in each row are as follows: 20, 22, 24…..**

**The number of seats in each row form an A.P. with the number of seats in first row (a) = 20.**

**Difference between number of seats in two successive rows (d) = 2**

**Total number of rows (n) = 27**

**Number of seats in 15th row (t₁₅) = ?**

**Total number of seats in 27 rows (S₂₇) = ?**

**tₙ = a + (n – 1) d**

**∴ t₁₅ = 20 + (15 – 1) × 2 **

**∴ t₁₅ = 20 + 14 × 2 **

**∴ t₁₅ = 20 + 28**

**∴ t₁₅ = 48**

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₂₇ = \(\large \frac {27}{2}\) [2 × 20 + (27 – 1) 2]**

**∴ S₂₇ = \(\large \frac {27}{2}\) [40 + 52]**

**∴ S₂₇ = \(\large \frac {27}{2}\) × 92**

**∴ S₂₇ = 1242**

**Ans:** There are 48 seats in the 15th row and 1242 seats in the auditorium.

**5. Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5° C more than sum of temperatures of Tuesday and Saturday. If the temperature of Wednesday was – 30° celsius then find the temperature on the other five days. **

**5. Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5° C more than sum of temperatures of Tuesday and Saturday. If the temperature of Wednesday was – 30° celsius then find the temperature on the other five days.****Solution:**

**Let the temperatures of Kargil from Monday to Saturday (in °C) which forms an A.P. be a – d, a, a + d, a + 2d, a + 3d, a + 4d respectively.**

**As per the given condition,**

**(a – d) + (a + 4d) = a + (a + 4d) + 5**

**∴ a – d + a + 4d = a + a + 4d + 5**

**∴ 2a + 3d = 2a + 4d + 5**

**∴ 3d = 4d + 5**

**∴ 4d – 3d = – 5**

**∴ d = – 5**

**New temperature on Wednesday is – 30°C … [Given]**

**∴ a + d = – 30**

**∴ a – 5 = – 30**

**∴ a = – 30 + 5**

**∴ a = – 25**

**Hence, **

**a – d = – 25 – (– 5) = – 25 + 5 = – 20**

**a = – 25**

**a + d = – 25 + (– 5) = – 25 – 5 = – 30**

**a + 2d = – 25 + 2(– 5) = – 25 – 10 = – 35**

**a + 3d = – 25 + 3(– 5) = – 25 – 15 = – 40**

**a + 4d = – 25 + 4(– 5) = – 25 – 20 = – 45**

** **

**Ans:** The temperatures from Monday to Saturday are – 20°C, – 25°C, – 30°C, – 35°C, – 40°C, – 45°C respectively.

**6. On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.**

**6. On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.****Solution:**

**The number of trees in each row upto the 25th row are as follows: 1, 2, 3, 4, …**

**The no. of trees planted in each row form an A.P. with no. of trees in first row (a) = 1**

**Difference between no. of trees planted in two successive rows (d) = 1**

**No. of rows (n) = 25**

**Total no. of trees planted (S₂₅) = ?**

** **

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₂₅ = \(\large \frac {25}{2}\) [2 × 1 + (25 – 1)1]**

**∴ S₂₅ = \(\large \frac {25}{2}\) [2 + 24]**

**∴ S₂₅ = \(\large \frac {25}{2}\) × 26**

**∴ S₂₅ = 25 × 13**

**∴ S₂₅ = 325**

** **

**Ans:** 325 trees were planted in 25 rows.

**Problem Set 3**

**Problem Set 3****1. Choose the correct alternative answer for each of the following sub questions.**

**1. Choose the correct alternative answer for each of the following sub questions.****(1) The sequence – 10, – 6, – 2, 2, … **

**(A) is an A.P., Reason d = – 16 **

**(B) is an A.P., Reason d = 4**

**(C) is an A.P., Reason d = – 4 **

**(D) is not an A.P.**

**Ans:** Option (B) : is an A.P., Reason d = 4

**(2) First four terms of an A.P. are ….., whose first term is – 2 and the common difference is – 2. **

**(A) – 2, 0, 2, 4 **

**(B) – 2, 4, – 8, 16**

**(C) – 2, – 4, – 6, – 8 **

**(D) – 2, – 4, – 8, – 16**

**Ans:** Option (C) : – 2, – 4, – 6, – 8

**(3) What is the sum of the first 30 natural numbers?**

**(A) 464 **

**(B) 465 **

**(C) 462 **

**(D) 461**

**Ans:** Option (B) : 465

**(4) For an given A.P. t₇ = 4, d = – 4 then a = … **

**(A) 6 **

**(B) 7 **

**(C) 20 **

**(D) 28**

**Ans:** Option (D) : 28

**(5) For a given A.P. a = 3.5, d = 0, n = 101, then tₙ = … **

**(A) 0 **

**(B) 3.5 **

**(C) 103.5 **

**(D) 104.5**

**Ans:** Option (B) : 3.5

**(6) In an A.P. first two terms are – 3, 4 then 21st term is … **

**(A) – 143 **

**(B) 143 **

**(C) 137 **

**(D) 17**

**Ans:** Option (C) : 137

**(7) If for any A.P. d = 5 then t₁₈ – t₁₃ = … **

**(A) 5 **

**(B) 20 **

**(C) 25 **

**(D) 30**

**Ans:** Option (C) : 25

**(8) Sum of first five multiples of 3 is… **

**(A) 45 **

**(B) 55 **

**(C) 15 **

**(D) 75**

**Ans:** Option (A) : 45

**(9) 15, 10, 5, … In this A.P. sum of first 10 terms is … **

**(A) – 75 **

**(B) – 125 **

**(C) 75 **

**(D) 125**

**Ans:** Option (A) : – 75

** **

**(10) In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = … **

**(A) 42 **

**(B) 38 **

**(C) 21 **

**(D) 19**

**Ans:** Option (B) : 38

**2. Find the fourth term from the end in an A.P. – 11, – 8, – 5, … , 49. **

**2. Find the fourth term from the end in an A.P. – 11, – 8, – 5, … , 49.****Solution:**

**For the reverse A.P. 49, …, – 5, – 8, – 11**

**a = 49, d = – 8 – (– 5) = – 8 + 5 = – 3**

**∴ Fourth term from the end of given A.P. is t₄ in the reverse A.P.**

**We know,**

**tₙ = a + (n – 1) d**

**∴ t₄ = 49 + (4 – 1) (– 3)**

**∴ t₄ = 49 + 3(– 3)**

**∴ t₄ = 49 – 9**

**∴ t₄ = 40**

**Ans:** Fourth term from the end of given A.P. is 40

**3. In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P. **

**3. In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.****Solution:**

**For an A.P., **

**t₁₀ = 46 and t₅ + t₇ = 52**

**We know that,**

**tₙ = a + (n – 1) d**

**∴ t₁₀ = a + (10 – 1) d**

**∴ 46 = a + 9d**

**∴ a + 9d = 46 …(i)**

**Now, **

**t₅ = a + (5 – 1) d**

**∴ t₅ = a + 4d and …(ii)**

**t₇ = a + (7 – 1) d**

**∴ t₇ = a + 6d …(iii)**

**We have,**

**t₅ + t₇ = 52 … [Given]**

**∴ a + 4d + a + 6d = 52 … [From (ii) and (iii)]**

**∴ 2a + 10d = 52**

**∴ a + 5d = 26 …. (iv) [Dividing both sides by 2]**

**Subtracting equation (iv) from equation (i)**

** a + 9d = 46**

** a + 5d = 26**

** (–) (–) (–)**

** _____________**

** 4d = 20**

**∴ d = \(\large \frac {20}{4}\)**

**∴ d = 5**

**Substituting d = 5 in equation (i) we get,**

**a + 9d = 46**

**∴ a + 9 × 5 = 46**

**∴ a + 45 = 46**

**∴ a = 46 – 45**

**∴ a = 1**

**Now,**

**t₁ = a = 1**

**t₂ = t₁ + d = 1 + 5 = 6**

**t₃ = t₂ + d = 6 + 5 = 11**

**t₄ = t₃ + d = 11 + 5 = 16**

**Ans:** The given A.P. is 1, 6, 11, 16…

**4. The A.P. in which 4th term is – 15 and 9th term is – 30. Find the sum of the first 10 numbers.**

**4. The A.P. in which 4th term is – 15 and 9th term is – 30. Find the sum of the first 10 numbers.****Solution:**

**Here, **

**t₄ = – 15, t₉ = – 30, S₁₀ = ?**

**tₙ = a + (n – 1) d**

**∴ t₄ = a + (4 – 1) d**

**∴ – 15 = a + 3d**

**∴ a + 3d = – 15 …(i)**

**Also, **

**t₉ = a + (9 – 1) d**

**∴ – 30 = a + 8d**

**∴ a + 8d = – 30 …(ii)**

**Subtracting (i) from (ii) we get,**

** a + 8d = – 30**

** a + 3d = – 15**

** (–) (–) (+)**

** ________________**

** 5d = – 15**

**d = \(\large \frac {15}{5}\)**

** ∴ d = – 3**

**Substituting d = – 3 in equation (i) we get,**

**a + 3d = – 15**

**∴ a + 3 (– 3) = – 15**

**∴ a – 9 = – 15**

**∴ a = – 15 + 9**

**∴ a = – 6**

**Now, **

**Sₙ = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**∴ S₁₀ = \(\large \frac {10}{2}\) [2 × (– 6) + (10 – 1) × (– 3)]**

**∴ S₁₀ = 5 [– 12 + 9 × – 3]**

**∴ S₁₀ = 5 [– 12 – 27]**

**∴ S₁₀ = 5 × – 39**

**∴ S₁₀ = – 195**

**Ans:** The sum of the first 10 terms of the A.P. is – 195.

**5. Two A.P.’s are given 9, 7, 5, … and 24, 21, 18, … . If the nth term of both the progressions are equal then find the value of n and nth term.**

**5. Two A.P.’s are given 9, 7, 5, … and 24, 21, 18, … . If the nth term of both the progressions are equal then find the value of n and nth term.****Solution:**

**The first A.P. is 9, 7, 5, …**

**Here, a = 9, d = 7 – 9 = – 2**

**tₙ = a + (n – 1) d**

**∴ tₙ = a + (n – 1) d**

**∴ tₙ = 9 + (n – 1) (– 2)**

**∴ tₙ = 9 – 2n +2**

**∴ tₙ = 11 – 2n …(i)**

**The second A.P. is 24, 21, 18….**

**Here, a = 24, d = 21 – 24 = – 3**

**tₙ = a + (n – 1) d**

**∴ tₙ = 24 + (n – 1) (– 3)**

**∴ tₙ = 24 – 3n + 3**

**∴ tₙ = 27 – 3n …(ii)**

**Now, **

**nth term of first A.P. = nth term of second A.P.**

**∴ 11 – 2n = 27 – 3n … [From (i) and (ii)]**

**∴ – 2n + 3n = 27 – 11**

**∴ n = 16**

**Now, **

**tₙ = 11 – 2n … [From (i)]**

**∴ tₙ = 11 – 2 × 16**

**∴ tₙ = 11 – 32**

**∴ tₙ = – 21**

**Ans:** The value of n is 16 and nth term is – 21.

**6. If the sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is – 3 then find the 10th term.**

**6. If the sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is – 3 then find the 10th term.****Given: **

**For an A.P.,**

**t₃ + t₈ = 7, **

**t₇ + t₁₄ = – 3, **

**To find: **

**t₁₀**

**Solution:**

**tₙ = a + (n – 1) d**

**∴ t₃ = a + (3 – 1) d**

**∴ t₃ = a + 2d …(i)**

**Also, **

**t₈ = a + (8 – 1) d**

**∴ t₈ = a + 7d …(ii)**

**Now, **

**t₃ + t₈ = 7 … [Given]**

**∴ a + 2d + a + 7d = 7 … [From (i) and (ii)]**

**∴ 2a + 9d = 7 …(iii)**

**tₙ = a + (n – 1) d**

**∴ t₇ = a + (7 – 1) d**

**∴ t₇ = a + 6d … (iv)**

**Also, **

**t₁₄ = a + (14 – 1) d**

**∴ t₁₄ = a + 13d …(v)**

**Now, **

**t₇ + t₁₄ = – 3 … [Given]**

**∴ a + 6d + a + 13d = –3 … [From (iv) and (v)]**

**∴ 2a + 19d = – 3 …(vi)**

**Subtracting (vi) from (iii) we get,**

** 2a + 9d = 7**

** 2a + 19d = – 3**

** (–) (–) (+)**

**________________**

** – 10d = 10**

**∴ d = \(\large \frac {10}{–\,10}\)**

**∴ d = – 1**

**Substituting d = – 1 in (iii), we get,**

**2a + 9d = 7**

**∴ 2a + 9(– 1) = 7**

**∴ 2a – 9 = 7**

**∴ 2a = 7 + 9**

**∴ 2a = 16**

**∴ a = \(\large \frac {16}{2}\)**

**∴ a = 8**

**tₙ = a + (n – 1) d**

**∴ t₁₀ = 8 + (10 – 1) (– 1)**

**∴ t₁₀ = 8 + (9) (– 1)**

**∴ t₁₀ = 8 – 9**

**∴ t₁₀ = – 1**

** **

**Ans:** The 10th term is – 1

**7. In an A.P. the first term is – 5 and last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference? **

**7. In an A.P. the first term is – 5 and last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?****Solution:**

**Here, **

**t₁ = a = – 5, tₙ = 45, Sₙ = 120, n = ?, d = ?**

**Now,**

**Sₙ = \(\large \frac {n}{2}\) [t₁ + tₙ]**

**∴ 120 = \(\large \frac {n}{2}\) [– 5 + 45]**

**∴ 240 = n [40]**

**∴ n = \(\large \frac {240}{40}\)**

**∴ n = 6**

**tₙ = a + (n – 1)d**

**∴ 45 = – 5 + (6 – 1)d**

**∴ 45 = – 5 + 5d**

**∴ 45 + 5 = 5d**

**∴ 5d = 50**

**∴ d = \(\large \frac {50}{5}\)**

**∴ d = 10**

**Ans:** There are 6 terms in the A.P and the common difference is 10.

**8. Sum of 1 to n natural numbers is 36, then find the value of n.**

**8. Sum of 1 to n natural numbers is 36, then find the value of n.****Solution:**

**The A.P. is 1, 2, 3, … , n.**

**They form an A.P. with t₁ = 1, d = 1, tₙ = n and Sₙ = 36**

**Now,**

**Sₙ = \(\large \frac {n}{2}\) [t₁ + tₙ]**

**∴ 36 = \(\large \frac {n}{2}\) [1 + n]**

**∴ 72 = n [1 + n]**

**∴ 72 = n + n²**

**∴ n² + n – 72 = 0**

**∴ n² + 9n – 8n – 72 = 0**

**∴ n (n + 9) – 8 (n + 9) = 0**

**∴ n + 9 = 0 or n – 8 = 0**

**∴ n = – 9 or n = 8**

** **

**But, n ≠ – 9 as n is a natural number.**

**∴ n = 8**

**Ans:** The value of n is 8.

**9. Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.**

**9. Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.****Solution:**

**Let the three parts of 207, which are in A.P. be a – d, a, a + d**

**Sum of three parts is 207 … [Given]**

**∴ a – d + a + a + d = 207**

**∴ 3a = 207**

**∴ a = \(\large \frac {207}{3}\)**

**∴ a = 69**

**Also, **

**The product of two smaller parts is 4623 … [Given]**

**∴ (a – d) × a = 4623**

**∴ (69 – d) × 69 = 4623**

**∴ 69 – d = \(\large \frac {4623}{69}\)**

**∴ 69 – d = 67**

**∴ 69 – 67 = d**

**∴ d = 2**

**Thus, the three parts of 207 are**

**a – d = 69 – 2 = 67, **

**a = 69**

**a + d = 69 + 2 = 71**

**Ans:** Three parts of 207 are 67, 69, 71.

**10. There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of the last three terms is 429. Write the A.P.**

**10. There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of the last three terms is 429. Write the A.P.****Solution:**

**Let the A.P. be a, a + d, a + 2d, a + 3d, …, a + 36d**

**Now, n = 37 (which is an odd number)**

**Middle term **

**= \(\large \frac {n\,+\,1}{2}\)**

**= \(\large \frac {37\,+\,1}{2}\)**

**= \(\large \frac {38}{2}\)**

**= 19th term.**

**∴ 18th, 19th, 20th terms are middle most terms.**

**The sum of the three middle most terms is 225 … [Given]**

**∴ t₁₈ + t₁₉ + t₂₀ = 225**

**∴ (a + 17d) + (a + 18d) + (a + 19d) = 225 … [∵ tₙ = a + (n – 1)d]**

**∴ 3a + 54d = 225**

**∴ a + 18d = 75 …(i) [Dividing both sides by 3]**

**Now, **

**35th, 36th, 37th terms are last three terms and their sum is 429.**

**∴ t₃₅ + t₃₆ + t₃₇ = 429**

**∴ (a + 34d) + (a + 35d) + (a + 36d) = 429**

**∴ 3a + 105d = 429**

**∴ a + 35d = 143 …(ii) [Dividing both sides by 3]**

**Subtracting (i) from (ii) we get,**

** a + 35d = 143**

** a + 18d = 75**

** (–) (–) (–)**

**_______________**

**17d = 68**

**∴ d = \(\large \frac {68}{17}\)**

**∴ d = 4**

**Substituting d = 4 in (i) we get,**

**a + 18d = 75**

**∴ a + 18 × 4 = 75**

**∴ a + 72 = 75**

**∴ a = 75 – 72**

**∴ a = 3**

**Ans:** The required A.P. is 3, 7, 11, 15…. 147

**11. If the first term of an A.P. is a, second term is b and last term is c, then show that the sum of all terms is \(\large \frac {(a\,+\,c)(c\,+\,b\,–\,2a)}{2(b\,–\,a)}\) **

**11. If the first term of an A.P. is a, second term is b and last term is c, then show that the sum of all terms is \(\large \frac {(a\,+\,c)(c\,+\,b\,–\,2a)}{2(b\,–\,a)}\)****Solution:**

**The A.P. is a, b, … , c**

**Here, **

**t₁ = a, d = b – a, tₙ = c**

**We know,**

**tₙ = a + (n – 1) d**

**∴ c = a + (n – 1) (b – a)**

**∴ c – a = (n – 1) (b – a)**

**∴ \(\large \frac {c\,–\,a}{b\,–\,a}\) = n – 1**

**∴ \(\large \frac {c\,–\,a}{b\,–\,a}\) + 1 = n**

**∴ \(\large \frac {c\,–\,a\,+\,b\,–\,a}{b\,–\,a}\) = n**

**∴ n = \(\large \frac {c\,+\,b\,–\,2a}{b\,–\,a}\)**

**We know,**

**Sₙ = \(\large \frac {n}{2}\) [t₁ + tₙ]**

**∴ Sₙ = \(\large \frac {c\,+\,b\,–\,2a}{2(b\,–\,a)}\) [a + c]**

**∴ Sₙ = \(\large \frac {(a\,+\,c)(c\,+\,b\,–\,2a)}{2(b\,–\,a)}\) **

**Hence proved.**

**12. If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)**

**12. If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)****Given: **

**S\(_p\) = S\(_q\)**

**To prove:**

**S\(_{p\,+\,q}\) = 0**

** **

**Proof: **

**Let a be the first term and d be the common difference of the A.P.**

**∴ S\(_n\) = \(\large \frac {n}{2}\) [2a + (n – 1) d]**

**The sum of first p terms of the A.P. is**

**S\(_p\) = \(\large \frac {p}{2}\) [2a + (p – 1) d] …(i)**

**The sum of first q terms of the A.P. is**

**S\(_q\) = \(\large \frac {q}{2}\) [2a + (q – 1) d] …(ii)**

**But S\(_p\) = S\(_q\)… [Given]**

**∴ \(\large \frac {p}{2}\) [2a + (p – 1) d] = \(\large \frac {q}{2}\) [2a + (q – 1) d] … [From (i) and (ii)]**

**∴ p[2a + pd – d] = q[2a + qd – d] … [Multiplying both sides by 2]**

**∴ 2ap + p²d – pd = 2aq + q2d – qd**

**∴ 2ap – 2aq + p²d – q²d – pd + qd = 0**

**∴ 2a(p – q) + (p² – q²)d – (p – q)d = 0**

**∴ 2a(p – q) + d (p + q) (p – q) – d(p – q) = 0 … [Using a² – b² = (a + b) (a – b)]**

**∴ (p – q) (2a + pd + qd – d) = 0**

**∴ 2a + (p + q – 1)d = \(\large \frac {0}{p\,–\,q}\)**

**∴ 2a + (p + q – 1)d = 0…(iii) [∵ p ≠ q, p – q ≠ 0]**

**Now,**

**S\(_{p\,+\,q}\) = \(\large \frac {p\,+\,q}{2}\) [2a + (p + q – 1) d]**

**∴ S\(_{p\,+\,q}\) = \(\large \frac {p\,+\,q}{2}\) × 0 … [From (iii)]**

**∴ S\(_{p\,+\,q}\) = 0**

**Hence, the sum of first (p + q) terms is zero.**

**13. If m times the mth term of an A.P. is equal to n times nth term then show that the (m + n)th term of the A.P. is zero.**

**13. If m times the mth term of an A.P. is equal to n times nth term then show that the (m + n)th term of the A.P. is zero.****Given: **

**mtₘ = ntₙ**

**To Prove: **

**t\(_{m\,+\,n}\) = 0**

**Proof:**

**Let a be the first term and d be the common difference of the A.P.**

**Then mth term of the A.P.**

**tₘ = a + (m – 1) d …(i)**

**and nth term of the A.P.**

**tₙ = a + (n – 1) d …(ii)**

**m(tₘ) = n(tₙ) … [Given]**

**∴ m [a + (m – 1)d] = n [a + (n – 1)d] …[From (i), (ii)]**

**∴ m[a + md – d] = n [a + nd – d]**

**∴ ma + m²d – md = na + n²d – nd**

**∴ ma – na + m²d – n²d – md + nd = 0**

**∴ (m – n) a + (m² – n²) d – d (m – n) = 0**

**∴ (m – n) a + (m + n) (m – n) d – d (m – n) = 0**

**∴ (m – n) [a + (m + n)d – d] **

**∴ a + (m + n) d – d = 0 … [∵ m ≠ n, m – n ≠ 0]**

**∴ a + (m + n – 1) d = 0 …(iii)**

**Now,**

**t\(_{m\,+\,n}\) = a + (m + n – 1) d**

**t\(_{m\,+\,n}\) = 0 …[From (iii)]**

**Hence, (m + n)th term of the A.P. is zero.**

**14. ₹1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find an interest amount after 20 years. For this complete the following activity.**

**14. ₹1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find an interest amount after 20 years. For this complete the following activity.****Simple interest = \(\large \frac {P\,×\,R\,×\,N}{100}\)**

**Simple interest after 1 year = \(\large \frac {1000\,×\,10\,×\,1}{100}\) = ___**

**Simple interest after 2 year = \(\large \frac {1000\,×\,10\,×\,2}{100}\) = ___**

**Simple interest after 3 year = \(\large \frac {□\,×\,□\,×\,□}{100}\) = 300**

**According to this the simple interest for 4, 5, 6 years will be 400, __, __ respectively.**

**From this d = __, and a = __**

**Amount of simple interest after 20 years**

**tₙ = a + (n – 1) d**

**t₂₀ = __ + (20 – 1) __**

**t₂₀ = __**

**Amount of simple interest after 20 years is = __**

**Solution:**

**Simple interest = \(\large \frac {P\,×\,R\,×\,N}{100}\)**

**Simple interest after 1 year = \(\large \frac {1000\,×\,10\,×\,1}{100}\) = 100**

**Simple interest after 2 year = \(\large \frac {1000\,×\,10\,×\,2}{100}\) = 200**

**Simple interest after 3 year = \(\large \frac {1000\,×\,10\,×\,3}{100}\) = 300**

**According to this the simple interest for 4, 5, 6 years will be 400, 500, 600 respectively.**

**From this d = 100, and a = 100**

**Amount of simple interest after 20 years**

**tₙ = a + (n – 1) d**

**t₂₀ = 100 + (20 – 1) 100**

**t₂₀ = 2000**

**Amount of simple interest after 20 years is = ₹2000**