Answer
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Hint: 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 \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
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 \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].