Question

# Find the sum of all positive integers, from $5$ to $1555$ inclusive, that are divisible by $5$A.$242489$B.$242580$C.$242420$D.$252420$

Hint : (Use the given series to find first term and common difference and proceed as considering the series an AP)

The first few terms of a sequence of positive integers divisible by $5$ is given by
$5,10,15,...$

The above sequence has a first term ${a_1} = 5$ and a common difference $d = 5$.

We need to know the rank of the term $1555$.

We use the formula for the ${n^{th}}$ term as follows :- ${a_n} = a + (n - 1)d$
Applying the formula

$1555 = {a_1} + (n - 1)d$

Substituting the values of ${a_{1\,}}$ and $d$ we get,

$1555 = 5 + 5(n - 1)$

Solve to obtain $n$
$n = 311$

We now know that $1555$ is the ${311^{th}}$ term, we can use the formula for the sum in AP .

We Know
${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$
We have to calculate ${S_{311}}$
${S_{311}} = \dfrac{{311}}{2}\left( {(2)5 + 310(5)} \right) = 242580 \\ \\$
Hence the Correct option is B.

Note :- In these types of questions we have to consider the given series as an AP, then solve it by using the formulas of nth term of AP and sum of N terms of an AP to get the result as done above.