Question

The last term of an arithmetic progression $2,{\text{ 5, 8, 11,}}....$ is ${\text{x}}$. The sum of the terms of the arithmetic progression is $155.$ Find the value of ${\text{x}}$.

Answer 1

Hint: In this question we have to find the value of ${\text{x}}$. For that we are going to find the value by using the given arithmetic progression. From the given arithmetic progression we are going to find the value of ${\text{n}}$ terms. By using ${\text{nth}}$ terms of arithmetic progression, we can get the expression of ${\text{x}}$. Next we are going to find the sum of ${\text{nth}}$ terms of arithmetic progression, then we get the value of ${\text{n}}$. Substitute the value of these in ${\text{x}}$, finally we get the value. Here given below in complete step by step solution.

Formula used: The general form of an arithmetic progression is ${\text{a, a + d, a + 2d, a + 3d}}$ and soon.
Thus ${{\text{n}}^{{\text{th}}}}$ term of an arithmetic progression series is \[{{\text{T}}_{\text{n}}}{\text{ = a + }}\left( {{\text{n - 1}}} \right){\text{d}}\] where \[{{\text{T}}_{\text{n}}} = {\text{ }}{{\text{n}}^{{\text{th}}}}\] and
${\text{a = }}$ first term. Here ${\text{d = }}$ common difference $ = {\text{ }}{{\text{T}}_{\text{n}}}{\text{ - }}{{\text{T}}_{{\text{n - 1}}}}$.
Sum of first ${\text{n}}$terms is ${{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left( {{\text{a + l}}} \right)$ where is the last term.

Complete step-by-step answer:
Given the last term of an arithmetic progression is $2,{\text{ 5, 8, 11,}}....$ is ${\text{x}}$.
And the sum of the terms of the arithmetic progression is $155.$
To find the value of ${\text{x}}$.
Here $2,{\text{ 5, 8, 11,}}....$ are in arithmetic progression with first term ${\text{a = 2}}$
And the common difference is $3$.
Let there be n terms in the arithmetic progression
${\text{x = }}{{\text{n}}^{{\text{th}}}}$ terms
${\text{x = a + }}\left( {{\text{n - 1}}} \right){\text{ d}}$
Substitute the values of terms,
${\text{x = 2 + (n - 1)3}}$
Multiple the values of terms,
${\text{x = 2 + 3n - 3}}$
Simplifying we get,
${\text{x = 3n - 1}}$
Now, $2{\text{ + 5 + 8 + 11 + }}....{\text{ + x = 155}}$
By using the sum of ${\text{n}}$terms of arithmetic progression,
${{\text{S}}_{\text{n}}} = \dfrac{{\text{n}}}{2}\left( {{\text{a + l}}} \right)$
Substitute the value of ${\text{a}}$,
$ \Rightarrow \dfrac{{\text{n}}}{2}\left( {2 + {\text{x}}} \right) = 155$
Substitute the value of ${\text{l}}$,
$ \Rightarrow \dfrac{{\text{n}}}{2}\left( {2 + 3{\text{n - 1}}} \right) = 155$
Simplifying we get,
$ \Rightarrow \dfrac{{\text{n}}}{2}\left( {{\text{3n - 1}}} \right) = 155$
Cross multiplication,
$ \Rightarrow {\text{n}}\left( {3{\text{n + 1}}} \right) = 155 \times 2$
Multiple the value of terms,
$ \Rightarrow {\text{n}}\left( {3{\text{n + 1}}} \right) = 310$
Multiply the term n into the brackets,
$ \Rightarrow 3{{\text{n}}^2} + {\text{n = 310}}$
Rearranging as an equation,
$ \Rightarrow 3{{\text{n}}^2} + {\text{n - 310 = 0}}$
Factorized the terms, we get
$ \Rightarrow 3{{\text{n}}^2}{\text{ - 30 n + 31 n - 310 = 0}}$
Splitting by middle term method,
$ \Rightarrow 3{\text{n}}\left( {{\text{n - 10}}} \right) + 31\left( {{\text{n}} - 10} \right) = 0$
Simplifying we get,
$ \Rightarrow \left( {{\text{n - 10}}} \right)\left( {3{\text{n + 31}}} \right) = 0$
Therefore, ${\text{n = 10}}$.
Now, n terms in the arithmetic progression
${\text{x = 3n - 1}}$
Substituting the value of n,
$ \Rightarrow {\text{x = 3 }} \times {\text{ 10 - 1}}$
Solve the terms,
$ \Rightarrow {\text{x = 30 - 1}}$
Subtracting we get,
${\text{x = 29}}$

The value of ${\text{x = 29}}$.

Note: We have used the terms in the given arithmetic progression that is,
In mathematics, an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15… is an arithmetic progression with a common difference of 2. A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.