Learn Exercise 5.2 with Free Lessons & Tips

(i)a = 7, d = 3, n = 8

= 7 + (8 − 1) 3

 = 28

Hence, a_n = 28

(ii) a = −18, n = 10, a_n = 0,

0 = − 18 + (10 − 1) d

Hence, common difference, d = 2

(iii)d = −3, n = 18, a_n = −5

−5 = a + (18 − 1) (−3)

−5 = a + (17) (−3)

a = 51 − 5 = 46

Hence, a = 46

(iv) a = −18.9, d = 2.5, a_n = 3.6, n = ?

3.6 = − 18.9 + (n − 1) 2.5

3.6 + 18.9 = (n − 1) 2.5

22.5 = (n − 1) 2.5

Hence, n = 10

(v) a = 3.5, d = 0, n = 105, a_n = ?

a_n = 3.5 + (105 − 1) 0

a_n = 3.5 + 104 × 0

a_n = 3.5

Hence, a_n = 3.5

a = 3

d = a₂ − a₁ = 15 − 3 = 12

a₅₄ = a + (54 − 1) d

= 3 + (53) (12) = 771

Let n^th term be 771.

 = a + (n − 1) d

771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

Therefore, 65^th term was 132 more than 54^th term.

(i)  a = 10

 d = a₂ − a₁ = 7 − 10 = −3

 = 10 + (30 − 1) (−3)

 = 10 + (29) (−3)

 = 10 − 87 = −77

Hence, the correct answer is C.

(ii) a = −3, d= 

Hence, the answer is B.

( i) first term = a = 7 ; common difference =d= 13-7=19-13=6;  n th term =tn=205 ; number of terms =n

Now, tn = a +(n-1).d

Or,  205  = 7 +(n-1).6, 

Or,  198 = (n-1). 6

Or,  33 = n-1. So, n = 34 (Answer).

( ii) first term  = a = 18, common difference =d= 15.5 -18= - 2.5; number of terms = n; n th term = tn= -47.

Now, tn = a +(n-1).d

Or, - 47 = 18 +(n-1).(-2.5)

Or,  - 65 = (n-1).(-2.5)

Or,  26 = (n-1). So, n = 27 (Answer)

First term  = a = 3; common difference  = d =8-3=13-8=5; number of terms  = n; n th term = tn = 78.

We know, tn = a + (n-1).d

So, 78 = 3 +(n-1).5

Or, 75 = (n-1).5, 

Or,  15 = n - 1, or, n =16 (Ans)

Let, first term = a; common difference = d; number of terms = n; n th term  = tn.

We know,  tn = a +(n-1).d

Putting, n=11 and n =16, we get, 

a + 10.d = 38 .... ( i); a + 15.d = 73 ..( ii)

By ( ii)  - ( i), 5.d = 35, or, d = 7.

Putting the value of d in ( i) , a+70=38, or, a = - 32.

So, 31st term = a +(n-1).d= -32 + (31-1).7 = - 32 + 210 = 178.

Given

 = 12, n = 50, a₅₀= 106

= a+2d= 12——(1)

a₅₀=a+49d= 106——(2)

(2)-(1)=> 47d= 94

d= 2

Substitute d=2 in (1)

a= 8

We have to find a_{29 =} 

                                  =64

First term  = a; common difference =d, number of terms=n; n th term = tn = a +(n-1).d.

Putting n=3 we get,    a+(3-1).d = 4 or,  a + 2.d = 4  ...( i)

Putting n=9,  a +(9-1).d = - 8 , 

a + 8d= - 8 ..( ii)

By ( ii) - ( i), 6.d = - 12, 

d = -2; 

Putting  d = -2 in ( i), a - 4 = 4, 

a=8.

Let nth term =0, 

a +(n-1).d =0,

 8 +(n-1).(-2)=0

n=5 

Therefore, the  term of the AP is zero.

First term =a; common difference =d; number of terms =n; n th term  tn= a +(n-1).d.

So, 17th term = a+(167-1).d = a+16d  ....( i); 10th term= a+(10-1).d = a + 9d ..( ii)

By (i) - ( ii), (a+16d) -(a+9d) =7 (given),  or, 7d = 7, or, d=1 (Ans)

Let the first terms of the two APs be a and b and the common difference of each be d. If number of terms = n and n th term is tn, then,  tn = a+(n-1).d and b+(n-1)d.

So, difference between the 100th terms = [ a+(100-1).d] -[b +(100-1)d] = (a+99d) -(b+99d) = 100 (given), or,  a- b = 100 ... ( i)

So, the difference between the 1000th terms =[a+(1000-1)d] - [b+(1000-1)d] = a - b =100 (Ans) [from  ( i)].

Ans : 100

The first 3 digit number divisible by 7 is 105 

  Last 3 digit number divisible by 7 is 994

formula- =a+(n-1)d

 Number of terms n = (l-a/d) +1

                                   =(994-105/7) +1

                                   =889/7 +1

                                  =127+1

                                  =128

Hence there are 128 three digit numbers divisible by 7.

Multiples of 4 between 10 and 250 are

12,16,20,24,.....248

Here first term a=12,common difference d=4,

Last term an=248, no.of terms n=? 

Wkt an=a+(n-1)d

248=12+(n-1)4

248=12+4n-4

248=8+4n

248-8=4n

240=4n

n=240/4

Therefore n=60

No.of multiples between 10 and 250 are 60.

a = 63

d = a₂ − a₁ = 65 − 63 = 2

n^th term of this A.P. 

= a + (n − 1) d

= 63 + (n − 1) 2

= 63 + 2n − 2

 = 61 + 2n................... (1)

3, 10, 17, …

a = 3

d = a₂ − a₁ = 10 − 3 = 7

n^th term of this A.P. = 3 + (n − 1) 7

 = 3 + 7n − 7

= 7n − 4 ..............(2)

It is given that, n^th term of these A.P.s are equal to each other.

Equating both these equations, we get

61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore, 13^th terms of both these A.P.s are equal to each other.

a₃ = 16

a + (3 − 1) d = 16

a + 2d = 16 ..............(1)

a₇ − a₅ = 12

[a+ (7 − 1) d] − [a + (5 − 1) d]= 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (1), we obtain

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be

4, 10, 16, 22, …

This A.P. can be written in reverse order as

253, 248, 243, …, 13, 8, 3

For this A.P.,

a = 253

d = 248 − 253 = −5

n = 20

a₂₀ = a + (20 − 1) d

= 253 + (19) (−5)

a₂₀₌ 158

Therefore, 20^th term from the last term is 158.

First term =a; common difference = d; number of terms = n; n th term = tn = a+(n-+).d.

So, sum of 4th and 8th terms  = a+(4-1).d  + a+(8-1).d = 2a+10d =24 (given) .... ( i)

Sum of 6th and 10th terms = a+(6-1)d + a+(10-1)d = 2a + 14d = 44 ....( ii)

By ( ii) - ( i),  4d = 20, or, d =5.

So, putting d in ( i),  2a+ 50 = 24, 

Or, 2a = - 26,  or, a = -13.

So, the first 3 terms of the AP are: a, a+d, a+2d = -13,-8,-3 ( Ans)

Subba Rao's salary is in the form of AP:

5000, 5200, 5400, …

Here, a = 5000

d = 200

Let after n^th year, his salary be Rs 7000.

Therefore, a_n = a + (n − 1) d

7000 = 5000 + (n − 1) 200

200(n − 1) = 2000

(n − 1) = 10

n = 11

Therefore, in 11th year, his salary will be Rs 7000.