Answer
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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.
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.
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