Question

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

Answer 1

Answer:

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,⋯⋯

Answer 2

$\LARGE{ \underline{\underline{ \bf{Required \: answer:}}}}$

The series of multiples of 3:

• 3, 6, 9, 12, 15......

Then the series of natural numbers that leaves remainder 1 when divided by 3 is:

• 4, 7, 10, 13.....

These numbers are in the form 3m + 1.

Here,

• First term = 4
• Common Difference = 3
• n = 40

By using formula,

$\large{ \boxed{ \rm{a_n = a + (n - 1)d}}}$

Plugging the given values:

⇒ a40 = 4 + (40 - 1)3

⇒ a40 = 4 + (39)3

⇒ a40 = 4 + 117

⇒ a40 = 121

Hence,

• The 40th term in the sequence is $121$

Explore more!!

• The general form of an Arithmetic Progression is a, a + d, a + 2d, a + 3d....

• Sum of first n terms of an AP: S =(n/2)[2a + (n- 1)d] or n/2(a + l)

• If a, b and c are three terms in AP then b = (a+c)/2. Or 2b = a + c.