**Q.1 In the given A.P: 26, 40, 54, 68 ………………. which term is 208?**

**Sol.**

From the given A.P:

a = 26,

d = 40 – 26 = 14,

a_{n} = 208,

n = ??

Since, a_{n }= a + (n – 1)d

Therefore, 208 = 26 + (n – 1)×14

14n = 208 + 14 – 26

Hence, n = 14

**14 ^{th} term of this given A.P is 208**

** **

**Q2. Find how many total numbers of terms are there in each of the following A.P **

**(i) 17, 26, 35, 44,……………………….179 **

**Sol**.

From the given A.P :

a = 17,

d = (26 – 17) = 9,

a_{n} = 179

n = ??

since, a_{n }= a + (n – 1)d

Therefore, 179 = 17 + (n – 1)×9

9n = 179 – 17 + 9

Therefore n = 19

**Hence, there are 19 terms in this given A.P**

(ii)

From the given A.P :

a =

d =

a_{n} =

n = ??

Since, a_{n }= a + (n – 1)d

Therefore,

21n=337+21-1

Therefore n= 17

**Hence, there are 17 terms in this given A.P**

** **

**Q3. Seventh term of an A.P is -1 and fourth term is 41. Determine the A.P and hence find its 17 ^{th} term.**

**Sol.**

Given, a_{7} = -1 and a_{4} = 41

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

-1 = a + (7 – 1)d

-1 = a + 6d . . . . . . . . . (1)

And 41 = a + (4 – 1)d

41 = a + 3d . . . . . . . (2)

Now, **equation (1) – equation (2)**

-1 – 41 = 6d – 3d Therefore, **d = -14**

Putting d = -14 in equation (2) we get

41 = a + 3×(-14)

Therefore, **a = 83**

**Therefore the required A.P is 83, 69, 55, 12, 41, 27, 13………**

Now, from the given A.P:

a = 83,

d = 69 – 83 = -14

n = 17

a_{17} = ??

Since, a_{n }= a + (n – 1)d

Therefore a_{17} = 83 + (17 – 1)× -14

a_{17} = 208 + 14 – 26

a_{17} = -141

**Therefore 17 ^{th} term of this given A.P is -141**

** **

**Q.4 There are 26 terms in an A.P of which fifth term is 61 and the last term is 292. Find 12 ^{th} term**

**Sol.**

Given,

a_{5} = a + (5-1)d = 61

61 = a + 4d…………………………………(1)

a_{26 }= a + (26 – 1)d = 292

292 = a + 25d……………………………(2)

**Equation (2) – Equation (1)**

292 – 61 = 25d – 4d

Therefore, **d = 11**

Putting the value of d in equation (1) we get

61 = a + 4 × 11

Therefore,** a = 17**

Hence a_{12 }= 17 + (12 – 1)×11 = 138

**Therefore 12 ^{th} term of this given A.P is 138**

** **

**Q.5 Find which term of an A.P is zero, if fourth term and tenth of an A.P are -26 and 52 respectively.**

**Sol.**

Given,

a_{n} = 0

a_{4} = a + (4 – 1)d = -26

-26 = a + 3d………………………………(1)

a_{10 }= a + (10 – 1)d = 52

52 = a + 9d……………………………(2)

**Equation (2) – Equation (1)** we get:

52 + 26 = 9d – 3d

Therefore, **d = 13**

On putting the values of d in equation (1) we get:

-26 = a + 3×13

Therefore, **a = -65**

Now, let n^{th} term be 0

Therefore, 0 = a + (n – 1)d

0 = -65 + (n – 1)×13

0 = -65 + 13n -13

n = 6

**Therefore 6 ^{th} term of the given A.P is zero.**

** **

**Q.6 Find the common difference d, if 14 ^{th} term of an A.P exceeds its 7^{th} term by 49.**

**Sol.**

Since, a_{n} = a + (n – 1)d

Therefore, a_{14 }= a + (14 – 1)d

Similarly, a_{7} = a + (7 – 1)d

Now, according to the given condition:

** a _{14} – a_{7 } = 49**

49 = {a + (14 – 1)d} – {a + (7 – 1)d}

49 = 13d – 6d

**Therefore, d = 7 (Common Difference) **

** **

**Q.7 Consider a series : 3, 20, 37, 54,71 …………………………..,Which term of this given A.P will be 289 more than its 32 ^{nd} term.**

**Sol.**

Given, a = 3

d = 20 – 3 = 17

Therefore, a_{32 }= a + (32 – 1)×d

a_{32} = 3 + 31×17

a_{32 }= 530

Now, according to the given condition:

** a _{n} – a_{32} = 289** (Since, a

_{n }> a

_{32})

(3+ (n – 1)17) – 530 = 289

819 = 3 + 17n – 17

n = 49

**Thus 49 ^{th} term of this given A.P will be 289 more than its 32^{nd} term.**

** **

**Q.8 Find the first three terms of an A.P, if sum of 12 ^{th} and 15^{th} term is 219 and sum of 5^{th} and 7^{th} term is 114.**

**Sol.**

Since, a_{n} = a + (n-1)d

Therefore, according to the given conditions:

**a _{12}+ a_{15 }= 219**

[a + (12-1)d] + [a + (15-1)d] = 219

2a + 25d = 219 ……………………………………………. (1)

And, a_{5 }+ a_{7} = 114

[a +(5-1)d] + [a+(7-1)d] = 114

2a + 10d = 114 ……………………………………………………(2)

Subtracting equation (2) from equation (1) we get

25d – 10d = 105

Therefore, **d = 7**

Putting values of d in equation (2) we get

2a + 10×7 = 114

Therefore **a = 22**

**Hence 1 ^{st} term of this A.P = a = 22**

** 2 ^{nd} term of this A.P = 22+(2-1)×7 = 29 **

** 3 ^{rd} term of this A.P = 22+(3-1)×7 = 36**

** **

**Q.9 ****Nikhil started his work in 2002 at an annual salary of Rs 9000 and each year he received an increment of Rs 300. Find in which year his income will reach Rs 12900?**

**Sol.**

Since, each year Nikhil’s salary is increased by Rs 300 and his starting salary was Rs 9000.

Therefore, this forms an A.P with a = 9000 and d = 300

Given, a_{n }= 12900 and n = ??

Since, a_{n} = a +(n-1)d

Therefore, 12900 = 9000 + (n-1)×300

300n = 12900 + 300 – 9000

Therefore **n = 14**

**Hence, the salary of Nikhil will be 12900 Rs after 14 years from 2002. **

** **

**Q.10 Find the first three terms of an A.P, if sum of 11 ^{th} and 14^{th} term is 126 and sum of 4^{th} and 6^{th} term is 66.**

**Sol.**

Since, a_{n} = a + (n-1)d

Therefore, according to the given conditions:

**a _{11}+ a_{14 }= 126**

[a + (11-1)d] + [a + (14-1)d] = 126

2a + 23d = 126 ……………………………………………. (1)

And, **a _{4 }+ a_{6} = 66**

[a +(4-1)d] + [a+(6-1)d] = 66

2a + 8d = 66 ……………………………………………………(2)

Subtracting equation (2) from equation (1) we get

23d – 8d = 60

Therefore, **d = 4**

Putting values of d in equation (2) we get

2a + 8×4 = 66

Therefore **a = 17**

Hence **1 ^{st} term of this A.P = a = 17**

** 2 ^{nd} term of this A.P = 17+(2-1)×4 = 21 **

** 3 ^{rd} term of this A.P = 17+(3-1)×4 = 25**

** **

**Q.11 In the first week of a particular year Deepak saved Rs 10 and then he increased his weekly savings by 2.25Rs. If Deepak was able to increase his weekly savings to Rs 46 in n ^{th} week. Find n**

**Sol. **

Money saved by Deepak in 1^{st} week = 10Rs

Money saved by Deepak in 2^{nd} week = 10 + 2.25 = 12.25Rs

Money saved by Deepak in 3^{rd} week = 12.25 + 2.25 =14.50Rs

Therefore, the above given situation forms an A.P: 10, 12.25, 14.50, 16.75…………..

Here, a = 10

d = 2.25

**a _{n }= 46**

Since, a_{n}= a+(n-1)d

Therefore 46= 10+(n-1)2.25

2.25n=46-10+2.25

Therefore **n= 17**

**Hence in 17 ^{th} week if that year Deepak was able to increase his weekly savings to Rs 46.**

** **

**Q.12 Find 26 ^{th} term from the last term of A.P: 7, 12, 17,22………..187.**

**Sol.**

Here, a = 187 (In this case, last term will be first term)

d= -5 (since the given A.P is reversed, therefore d = 12 -17 = 7 – 12 = -5)

And, n = 26

Since, a_{n}= a + (n – 1)d

Therefore a_{n} = 187 + (26 – 1)×-5

** a _{n}= 62**

**Therefore 26 ^{th} term from the last term is 62.**

** **

**Q.13 Consider two A.Ps :- 9, 15, 21, 27,…….. and 63,66, 69, 72,…….. Find the value of n, such that the n ^{th} term of both the A.P’s is equal.**

**Sol.**

For First A.P, a = 9 and d = 15-9 = 6

**Since a _{n} = a+(n-1)d**

Therefore a_{n}= 9+(n-1)6

(Or) **a _{n}= 6n+3**

For second A.P, a = 63 and d = 66-63 = 3

Since, a_{n} = a + (n – 1)d

Therefore, a_{n} = 63 + (n – 1)×3

(Or) **a _{n }= 3n +60**

Now, according to the given condition:

3n+60=6n+3

Therefore **n = 19**

**Therefore, 19 ^{th} term of both the A.P’s is equal.**

**Q.14 Find how many three digit numbers are divisible by 9**

**Sol.**

First 3- digit number divisible by 9 = 108

Second 3- digit number divisible by 9 = 117

Similarly last 3-digit number divisible by 9 = 999

Hence it forms an A.P with **d = 9**,

108, 117, 126……………999

Now, from the above A.P:

a= 108, d= 9, a_{n}= 999

n=??

Since, a_{n}= a+ (n-1)×d

Therefore 999= 108+ (n-1)×9

9n=999-108+9

Therefore **n= 900**

**Hence, there are 900 three digit numbers that are divisible by 9.**

** **

** Formula for Sum of first n terms:
**

**Let us suppose that there are n terms in an A.P, therefore sum of first n terms of an A.P is given by S _{n} = n2 [2a+(n-1)d]**

**Here, a = first term, d = common difference and n = number of terms whose sum is to be determined.**

**REMEMBER: a _{n }= S_{n }– S_{n–1} **

**Example-1: In an A.P -2, 5, 12, 19,……… Find the sum of first 8 terms.**

**Sol.**

Here, a = -2 and d = (12-5) = 7 and n = 8

Since, **S _{n} = n2 [2a+(n-1)d]**

Therefore S_{n} =

= 4 × [45]

**Therefore sum of first 8 terms = 180.**

** **

**Example – 2 How many terms of an A.P 34, 43, 52, 61,………………… must be taken so that there sum is equal to 1002.**

**Sol.**

Given, a = 34, d = 9 and S_{n} = 1002

Since, **S _{n} = n2 [2a+(n-1)d]**

Therefore, S_{n} = 1002 =

1002×2 = n×[59 + 9n]

**9n ^{2} + 59n – 2004 = 0**

Now, from the above quadratic equation: a = 9, b = 59 and c = -2004

Substituting the values of a, b and c in **quadratic formulae** we get:

n =

** n=12 (neglecting the negative terms)
**

**Therefore, the sum of first 12 terms in the given A.P will be 2004.**

** **

**Example-3 Find the sum of first 100 odd positive integers.**

**Sol.**

Odd positive integers can be written as: **3, 5, 7, 9, 11……**

Here a = 3, d = 2 and n = 100

Since ** ****S _{n} = n2 [2a+(n-1)d]**

Therefore, S_{n} =

= 50[6+198]

= 10200

**Therefore the sum of first 100 odd positive integers = 10200**

** **

**Example – 4 The n ^{th} term of any particular series in given by (3n – 5). Find sum of first 16 terms.**

**Sol. **

Given, **a _{n} = 3n – 5**

Therefore, a_{1} = 3×1 – 5 = -2

a_{2} = 3×2 – 5 = 1

a_{3} = 3×3 – 5 = 4

Therefore, **the series becomes -2, 1, 4 . . . . .**

Since [1 – (-2)] = (4 – 1) = 3 = d

Hence, the above series forms an A.P with a = -2 and d = 3

Since, the sum of first n terms is given by: S_{n} =

S_{n} =

= 8(-4 + 45)

= 328

**Therefore the sum of first 16 terms is 328**