Step-by-step explanation:

1) given,

3,6,9 ate in A.P

first term(a)=3

difference (d)=3

10 th term is 3 +9(3)

=30

so 10 th term is 30

2)given,

n th term of A P

IS given by 3n+2

6th term is ?

put n =6

then =3(6)+2

=20

so 6th term of A.P is 20

3)given ,

2,x,14 are in AP

X is ?

(2+14)/2=x

16=2x

x is 8

4)given,

first term of AP (a)=5

difference in A.P(d)=7

8 th term is a+7d

5+7(7)=54

so 8 th term in AP is 54

5)given,

first three terms in AP

√2 2√2 3√2

first term is √2

common difference is √2

4 th term is a+3d

√2+3(√2)

4√2

so 4 th term is 4√2

6)given,

5(5 th term)= 10(10 th term)

5(a+4d) =10(a+9d)

5a+20d= 10a +90d

-5(a)=70d

a= -14(d)

a+14d=0

so 15 th term of A.P is 0

7)given,

no of 3 digit numbers divisible by 5

numbers are 100,995

first term of AP IS 100

let term n be 995

difference is 5

a n =100+(n-1)d

995=100+(n-1)5

5n-5=895

5n =900

so n is 180

so 180 3 digit numbers are divisible by 5

8)given ,

a=8

d=2

10 th term of AP

8+9(2)

=26

so 10 th term of AP IS 26

9)given,

2 n d term of AP IS 13->(1)

5 TH term of AP IS 25->(2)

a+d=13

a+4d=25

3d is 12

d is 3

a is 13-3

a is 10

7 th term is 10+6(3)=28

10) given,

a is 3

d is 12

54 th term is 3+53(12)=639

let the term be n

3+(n-1)12=639+144

3+12n-12=783

12n-9 is 783

n is 792/12

66=n

so the 66 th term is 144 +54 th term