Question

Solve the equation $2 + 5 + 8 + \ldots + x = 155$.

Answer 1

Here, we need to solve the given equation. The left hand side of the given expression forms an A.P. We will solve the equation using the formula for ${n^{{\rm{th}}}}$ term of an A.P. to find the number of terms in the A.P. Then, we will use the formula for the sum of $n$ terms of an A.P. to find the required value of $x$.

Formula Used:
We will use the following formulas:
1.The ${n^{{\rm{th}}}}$ term of an A.P. is given by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the A.P. and $d$ is the common difference.
2.The sum of $n$ terms of an A.P. is given by the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ is the first term of the A.P. and $d$ is the common difference.

Complete step-by-step answer:
We will solve the equation using the formula for ${n^{{\rm{th}}}}$ term of an A.P. and the formula for the sum of $n$ terms of an A.P.
The left hand side of the given expression is $2 + 5 + 8 + \ldots + x$.
We can observe that the series $2 + 5 + 8 + \ldots + x$ is the sum of an A.P., where the first term is 2, the common difference is $5 - 2 = 3$, and the last term is $x$.
Let the number of terms in the sum $2 + 5 + 8 + \ldots + x$ be $n$.
Substituting $a = 2$ and $d = 3$ in the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, we get
${S_n} = \dfrac{n}{2}\left[ {2\left( 2 \right) + \left( {n - 1} \right)3} \right]$
It is given that the sum $2 + 5 + 8 + \ldots + x$ is equal to 155.
Therefore, we get
$\Rightarrow 155 = \dfrac{n}{2}\left[ {2\left( 2 \right) + \left( {n - 1} \right)3} \right]$
Multiplying the terms in the expression using the distributive law of multiplication, we get
$\Rightarrow 155 = \dfrac{n}{2}\left[ {4 + 3n - 3} \right]$
Simplifying the expression, we get
$\Rightarrow 155 = \dfrac{n}{2}\left[ {1 + 3n} \right]$
Multiplying both sides of the expression by 2, we get
$\Rightarrow 310 = n\left[ {1 + 3n} \right]$
Multiplying the terms in the expression using the distributive law of multiplication, we get
$\Rightarrow 310 = n + 3{n^2}$
Rewriting the expression, we get the equation
$\Rightarrow 3{n^2} + n - 310 = 0$
Here, we get a quadratic equation. We will split the middle term and factorise to solve the quadratic equation.
Factoring the quadratic equation, we get
$\begin{array}{l} \Rightarrow 3{n^2} + 31n - 30n - 310 = 0\\\ \Rightarrow n\left( {3n + 31} \right) - 10\left( {3n + 31} \right) = 0\\\ \Rightarrow \left( {3n + 31} \right)\left( {n - 10} \right) = 0\end{array}$
Therefore, either $3n + 31 = 0$ or $n - 10 = 0$.
Simplifying the expressions, we get
$n = - \dfrac{{31}}{3}$ or $n = 10$.
The value of $n$ cannot be negative because the number of terms in the A.P. cannot be negative.
Therefore, we get
$n = 10$
Now, we know that $x$ is the last term of the A.P.
This means that $x$ is the 10th term of the A.P.
Substituting ${a_n} = x$, $a = 2$, $d = 3$, and $n = 10$ in the formula ${a_n} = a + \left( {n - 1} \right)d$, we get
$x = 2 + \left( {10 - 1} \right)3$
Subtracting the terms in the parentheses, we get
$\Rightarrow x = 2 + \left( 9 \right)3$
Multiplying the terms in the parentheses, we get
$\Rightarrow x = 2 + 27$
Adding the terms in the parentheses, we get
$\Rightarrow x = 29$
Therefore, we get the value of $x$ as 29.

Note: The series given in the question is in Arithmetic Progression. Arithmetic progression is a series or sequence in which the consecutive terms differ by a common difference. We used the term “quadratic equation” in our solution. A quadratic equation is an equation of degree 2. It is of the form $a{x^2} + bx + c = 0$, where $a$ is not equal to 0. A quadratic equation has 2 solutions.
We have used the distributive law of multiplication in the solution. The distributive law of multiplication states that $a\left( {b + c} \right) = a \cdot b + a \cdot c$.