Question

Find the sequence of natural numbers which lives remained 1 on divisible by 3. Find it's 40th term. ​

Answer 1

Answer:

The three digit numbers which are divisible by 9 are

108,117,126,...……..999

which forms an A.P

the first term of this A.P is a

1

=108

second term of this A.P is a

2

=117

common difference of this A.P is

d=a

2

−a

1

=117−108=9

the nth term of this A.P is given by

a

n

=a

1

=(n−1)d

⟹a

n

=108+(n−1)9

⟹a

n

=108+9n−9

⟹a

n

=99+9n......eq(1)

since last term of this A.P is a

n

=999

hence for finding the number of terms (n) in this A.P put a

n

=999 in eq(1)

⟹999=99+9n

⟹9n=999−99

⟹9n=900

⟹n=

9

900

⟹n=100

hence there are 100 three digit terms which are divisible by 9.

Answer 2

Step-by-step explanation:

The sequence must be in the form of 3k+1

So the sequence is 1,4,7,10,....

The sequence must be in the form of 3k+2

So the sequence is 2,5,8,11,⋯⋯

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