## Chapter – 5: Arithmetic Progressions

**Que 1.** How many two digits numbers are divisible by 3?

**Solution:** First two digit number divisible by 3 = 12

Last two digit number divisible by 3 = 99

An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.

A.P = 12,15,18,…,99

here

First term (a) = 12

Common difference (d) = 3

Let us consider there are n numbers then

a_{n} = 99

a + (n – 1)d = 99

12 + (n – 1)3 = 99

12 + 3n – 3 = 99

n = 29+1

n = 30

**∴ Two digit numbers divisible by 3 = 30.**

**Que 2.** In an AP, if the common difference (d) = -4 and the seventh term (a_{7}) is 4, then find the the first term.

**Solution: **we know,a_{n }= a + (n−1)d

so, a_{7 }= a + (7−1)d

or, 4 = a + 6 (−4)

or, 24 + 4 = a

∴ a = 28

**Que 3. **Which term of AP 8,14,20,26…will be 72 more than its 41st term?

**Solution: **

**Given : **

8, 14, 20, 26, ..

first term , a = 8 , common difference , d = 14 – 8 = 6

nth term = 72 + a41

a + (n -1)d = 72 + a41

**[nth term = a + (n -1)d]**

8 + (n – 1) 6 = 72 + a + (41 – 1) d

8 + 6n – 6 = 72 + 8 + 40 × 6

2 + 6n = 72 + 8 + 240

2 + 6n = 80 + 240

2 + 6n = 320

6n = 320 – 2

6n = 318

n = 318/6

**n = 53**

**Hence, 53rd term of the A.P 8, 14, 20, 26, … will be 72 more than its 41st term.**

**Que 4. **Write the nth term of A.P. 1/m, (1+m)/m, (1+2m)/m, ……..?

**Solution:** We know that,

**Que 5. **Find the middle term of 6, 13, 20… 216?

**Solution:** The given AP is 6,13,20,……..,216.

First term, a = 6

Common difference d = 13-6 = 7

Suppose these are n terms in the given AP. Then,

Hence, the middle term of the given AP is 111.

**Que 6. **The 8th term of AP is zero. Prove that 38th term is triple of 18th term.

**Solution:** T_{8 }= a + 7d = 0

⇒ a = −7d

T_{38} = a + 37d

= −7d + 37d

= 30d

T_{18} = a + 17d

= −7d + 17d

= 10d.

Clearly T_{38}=3(T_{18}).

**Que 7. **Find the sum of all two digit positive odd numbers?

**Solution:** All two-digit odd positive numbers are 11, 13, 15, 17, …., 99. which are in AP with a = 11, d = 2, l = 99

Let the number of terms be n.

a_{n }= 99

a + (n − 1)d = 99

11 + (n − 1) × 2 = 99

n = 45

Sum of n terms is given by

S_{n} = n/2(a + l)

S_{n} = 45/2(11 + 99) = 2475

Therefore, the sum of all two-digit odd positive numbers is 2475.

**Que 8. **Three numbers are in AP and their sum is 21, find the middle number ?

**Solution:** Let the numbers be,

a − d, a, a + d

Given,

a − d + a + a + d = 21

3a = 21

∴ a = 7

Hence the middle number.

**Que 9. **If 7 times the 7th term of an AP is equal to 11 times its 11th term. Then find its 18th term.

**Solution:** Let a and d be first term and common difference respectively

According to the question,

7 times the 7th term of an AP is equal to 11 times its 11th term

⇒ 7a7 = 11a11

As we know, nth term of an AP is

an = a + (n – 1)d

where a = first term

an is nth term

d is the common difference

we have,

7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a + 68d = 0

⇒ a + 17d = 0

⇒ a18 = 0

18th term of AP is 0

**Que 10. **The consecutive terms of an AP are 2, x, 26, find the value of x?

**Solution: **x = 14

Step-by-step explanation:

2, x, 26 are in AP.

Hence,

x – 2 = 26 – x

2x = 28

x = 14

**Que 11. **For what value of p is 2p+1, 13, 5p-3, are 3 consecutive terms of an AP?

**Solution:** Given 2p+1, 13 , 5p-3 are in AP.

To find: value of P

If a,b,c are in AP then we know that

2b = ( a+c)

Here a = 2p + 1

b = 13

c = 5p – 3

2 (13) = (2p+1+5p-3)

26 = 7p-2

26 + 2 = 7p

28 = 7p

p = 28/7 = 4

Therefore the value of P is 4.

**Que 12. **Which term of the AP: 3, 8, 13, 18… is 78?

**Solution:** The formula for nth term of an AP is aₙ = a + (n – 1) d

The given arithmetic progression is 3, 8, 13, 18, …

First term: a = 3

Second term: a + d = 8

Common difference: d = 8 – 3 = 5

a_{n} = a + (n – 1) d

a_{n} = 78, n = ?

3 + (n – 1) 5 = 78

5(n – 1) = 78 – 3

n – 1 = 15

n = 16

78 is the 16^{th} term of the given AP.

**Que 13. **Write the 5th term from the end of the AP 3,5,7,9 .. 201?

**Solution:**

Given :

A.P. 3, 5, 7, 9, ….,201.

**Let ‘l’ be the last term of an A.P**

Here, l = 201, a = 3, a2 = 5

d = a2 – a

d = 5 – 3 =** 2**

nth term from the end = l – (n – 1)d

5th term from the end = l – (n – 1)d

5th term from the end = 201 – (5 – 1) × 2

5th term from the end = 201 – 4 × 2

5th term from the end = 201 – 8

5th term from the end =** 193**

Hence, the 5th term from the end of the A.P is 193.

**Que 14. **From the given AP: 8, 10, 12… Find the sum of its last 10 terms if it has 60 terms?

**Solution: **A.P is 8,10,12………………….

Then first term is 8 and common difference is 2.

a_{n }= a + (n−1)d.

Then a_{60 }= 8 + (60−1)2 = 8 + 59 × 2 = 126

Last 10 term =a_{51}, a_{52}………………a_{60}

Then a_{51 }= 8 + (51 − 1)2 = 108

So total of last 10 terms = 10/2 [108 × 2 + (10 − 1)2] = 216 + 81 = 234 × 5 = 1170

**Que 15. **Find the number of terms of an AP 5, 9, 13 …185.

**Solution:**

**Given:**

l = 185

a = 5

d = 4

**To find:**

n = ?

**Solution:**

l = a + (n – 1)d

185 = 5 + (n – 1)4

185 = 5 + 4n -4

185 = 1 + 4n

185 – 1 = 4n

184 = 4n

184/4 = n

46 = n