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Question

The sum of the numbers between 17 and 520 that are divisible by 6 is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper PDF (31-Jan-2021) (Shift 1)
The correct answer is
22428

Identifying the Sequence Terms

We need to find the sum of numbers between 17 and 520 that are divisible by 6. This forms an arithmetic progression.

  • First term ($a$): The smallest number greater than 17 that is divisible by 6 is 18. So, $a = 18$.
  • Last term ($l$): The largest number less than 520 that is divisible by 6. Divide 520 by 6: $520 \div 6 \approx 86.67$. Multiply the integer part by 6: $86 \times 6 = 516$. So, $l = 516$.
  • Common difference ($d$): Since we are looking for numbers divisible by 6, the common difference is $d = 6$.

Calculating the Number of Terms

Use the formula for the $n$-th term of an arithmetic progression: $l = a + (n-1)d$.

Substitute the values:

$516 = 18 + (n-1)6$

Subtract 18 from both sides:

$516 - 18 = (n-1)6$

$498 = (n-1)6$

Divide by 6:

$498 \div 6 = n-1$

$83 = n-1$

Add 1 to find $n$:

$n = 83 + 1 = 84$. There are 84 numbers between 17 and 520 divisible by 6.

Calculating the Sum of the Sequence

Use the formula for the sum of an arithmetic progression: $S_n = \frac{n}{2}(a + l)$.

Substitute the values of $n$, $a$, and $l$:

$S_{84} = \frac{84}{2}(18 + 516)$

$S_{84} = 42(534)$

Calculate the final sum:

$S_{84} = 22428$.

The sum of the numbers between 17 and 520 that are divisible by 6 is 22428.

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