Question

1) The 5th term of an AP is 22 and its 9th term is six times the 2nd term. Find that AP.

2) Find the number of all the natural numberslying between 101 and 999 which are divisible by 2 as well as 5.​

Answer 1

Answer:

1) AP-2,7,12,17,22...

2) 89

Step-by-step explanation:

1)

a+4d=22    ----(1)

now

a+8d=6(a+d)

a+8d=6a+6d

5a=2d

$a=\frac{2d}{5\\}$   ---(2)

Substituting (2) in (1)

$\frac{2d}{5}$+4d=22

Multipling by 5

2d+20d=110

22d=110

d=5  ---(3)

Substituting  (3) IN (2)

a=2

therefore AP is 2,7,12,17,22...

2) here the given numbers should be multiples of 10 thereby can be divisible by 2 and 5 both

AP=110,120,130...990

a=110

d=10

l=990

here we take l=$a_{n}$

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

990=110+(n-1)10

880=10n-10

890=10n

n=$\frac{890}{10}$

n=89

Therefore 89 numbers can be divisible by 2 as well as 5.

Pl. give brainliest dude