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**Hint:**Start with finding the first term and the common difference and then use the formula $S = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ to find the number of terms in the given series. The expression for the nth term is $a + \left( {n - 1} \right)d$ , so substitute the values in it to find the required value of ‘x’.

**Complete step-by-step answer:**Here in this problem, we are given a finite series, i.e. $1 + 4 + 7 + ...... + x$ and the sum of this series is given to be $287$ . With this information, we need to find the last term of the series, which is defined as $x$ . And we have four options for choosing the correct answer.

For the series $1 + 4 + 7 + ...... + x$ , the first term is $1$ and the last term is $x$ .

The difference can be found as:

$ \Rightarrow {\text{ Second term}} - {\text{Third term}} = 4 - 1 = 3$ and also ${\text{Fourth term}} - {\text{Third term}} = 7 - 4 = 3$

The common difference will be $3$ .

So, the series is having the same difference between all the terms, therefore, it is an Arithmetic Progression.

If all the terms of a progression except the first one exceeds the preceding term by a fixed number, then the progression is called arithmetic progression. If a is the first term of a finite AP and d is a common difference, then AP is written as $a,a + d,a + 2d,....,a + \left( {n - 1} \right)d$ .

The sum of $'n'$ terms of AP is the sum (addition) of the first $'n'$ terms of the arithmetic sequence. It is equal to n divided by $2$ times the sum of twice the first term - ‘a’ and the product of the difference between second and first term ‘d’ also known as common difference, and $\left( {n - 1} \right)$ , where n is the numbers of terms to be added.

$ \Rightarrow $ Sum of the first $'n'$ term $ = \dfrac{n}{2}\left( {{1^{st}}term + {n^{th}}term} \right) = \dfrac{n}{2}\left( {a + a + \left( {n - 1} \right)d} \right) = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$

For the given AP: $1 + 4 + 7 + ...... + x = 287$ , we write it as:

$ \Rightarrow $ Sum of $'n'$ the term $ = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) \Rightarrow 287 = \dfrac{n}{2}\left( {2 \times 1 + 3 \times \left( {n - 1} \right)} \right)$

Therefore, we get an equation with one unknown, i.e. $'n'$ . Let’s solve it to find the value of the number of terms.

$ \Rightarrow 287 = \dfrac{n}{2}\left( {2 + 3n - 3} \right) = \dfrac{n}{2}\left( {3n - 1} \right)$

On transposing $2$ to another side, we get:

$ \Rightarrow 287 \times 2 = 3n \times n - n \Rightarrow 3{n^2} - n - 574 = 0$

Now we can use the Quadratic formula to solve this equation, i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for an equation $a{x^2} + bx + c = 0$

$ \Rightarrow 3{n^2} - n - 574 = 0 \Rightarrow n = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 3 \times \left( { - 574} \right)} }}{{2 \times 3}} = \dfrac{{1 \pm \sqrt {1 + 12 \times 574} }}{6}$

We can solve the radical sign first as:

$ \Rightarrow n = \dfrac{{1 \pm \sqrt {1 + 12 \times 574} }}{6} = \dfrac{{1 \pm \sqrt {1 + 6888} }}{6} = \dfrac{{1 \pm \sqrt {6889} }}{6}$

As we know that $\sqrt {6889} = 83$ , let’s substitute it in the above equation:

$ \Rightarrow n = \dfrac{{1 \pm \sqrt {6889} }}{6} = \dfrac{{1 \pm 83}}{6} = \dfrac{{84}}{6} or \dfrac{{ - 82}}{6} = 14{\text{ or }} - 13.66$

Since ‘n’ is the number of terms and can never be negative or fraction.

Therefore, we have $n = 14$

The last term can be defined as $a + \left( {n - 1} \right)d$ and in this series last term is ‘x’

$ \Rightarrow x = a + (n - 1)d$

Now let’s substitute the values in it, we get:

$ \Rightarrow x = a + (n - 1)d = 1 + \left( {14 - 1} \right) \times 3 = 1 + 13 \times 3 = 1 + 39 = 40$

Thus, we get the required as $x = 40$

**Hence, the option (B) is the correct answer.**

**Note:**In a question like this, the use of properties of AP is very useful. We can check whether the given series is in AP or GP using their common difference and common ratio. If the successive terms are obtained by multiplying a constant to the term, then the series is known as Geometric Progression (GP).

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