Question

Write down the sequence of natural numbers
which leave a remainder 3 on division by
6. What is the 10th term of this sequence .
How many terms of this sequence are
between two and 400.​

Answer 1

Answer:

66 terms

Step-by-step explanation:

n = qd + r, where n is the number, q is quotient, d is divisor and r is remainder.

Here, let any term of this seq. be n.

n = required term ; quotient = q

divisor = 6, r = 3

So, term = 6q + 3

General term of this AP is 6q + 3. Therefore, 10th term is:

10th term = 6(10) + 3 = 63

Let 400 be xth term(nearest) of the seq.

=> 6x + 3 = 400

=> 6x = 396

=> x = 66

Since 2 is even lesser that the 1st term in the seq., value satisfying 6x(<2) + 3 doesnt matter.

Hence, there are 66 terms.

Answer 2

$\sf{\huge{\underline{Answer:)-   }}}$

As we know that,

• $\sf{n=qd+r   }$

where,

• $\sf{q=quotient    }$

• $\sf{ n=the no of terms   }$

• $\sf{r=reminder   }$

• $\sf{ d=divisor   }$

According to the question ,

• The divisor(d)=6
• reminder(r)=3

LET,the quotient =q

soo,

• term=6q+3

so,The G.F of the A.P =6q+3

The 10th term=

• 6q+3
• 6(10)+3
• 60+3
• 63

Let's assume the xth term of the sequence is 400.

HENCE,

• $\sf{ 6x+3=400  }$

• $\sf{6x=396   }$

• $\sf{  x=66 }$

$\therefore$There are 66 terms of the sequence between 2 and 400

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