Question

Find the value of $x$ such that $1 + 4 + 7 + 10 + ..... + x = 715$ .

Answer 1

In the above question we have given a sequence in which we have to find the value of $x$. In order to solve this question firstly you have to check whether the given sequence is A.P or not. Then apply the concept of Arithmetic Progression. Also remind all the formulas for finding the sum of $n$ terms and the ${n^{th}}$ term of an A.P.

Formula used:
The formula for sum of $n$ terms of an A.P is given by
${S_n} = \dfrac{n}{2} \times [2a + (n - 1)d]$
Where, $n$ is the total number of terms
$a$ is the first term
$d$ is the common difference
The formula for finding ${n^{th}}$ term of an A.P is given by
${a_n} = a + (n - 1)d$
Where, $n$ is the total number of terms
$a$ is the first term
$d$ is the common difference

Complete step-by-step answer:
Here in the question the sequence is given as
$1 + 4 + 7 + 10 + ..... + x = 715$
Firstly, check whether it is an AP or not. For this you have to find the difference between the first and second term and if it is equal to the difference of second and third then it is an A.P.
${a_2} - {a_1} = 4 - 1 = 3$
${a_3} - {a_2} = 7 - 4 = 3$
As we can see that both the differences are equal so, it is an A.P.
Hence, we get
$a = 1$
And, $d = 4 - 1$
$\Rightarrow d = 3$
Also the sum of all the terms in the AP is given in the question as
${S_n} = 715$
Let us assume that the given A.P contains $n$ terms
As we know that the sum of $n$ terms of an A.P is given by
${S_n} = \dfrac{n}{2} \times [2a + (n - 1)d]$
On putting all the values we have
$715 = \dfrac{n}{2} \times [2 \times 1 + (n - 1) \times 3]$
$\Rightarrow n(3n - 1) = 1430$
On further solving we will get the equation as
$\Rightarrow 3{n^2} - n - 1430 = 0$
Now solve this quadratic equation by using factoring method
First splitting middle term,
$\Rightarrow 3{n^2} - 66n + 65n - 1430 = 0$
Now on factorising we get,
$\Rightarrow (n - 22)(3n + 65) = 0$
$\Rightarrow n = 22$ {neglecting second factor as it cannot be negative}
Now here $x$ is the last term of the given A.P. and the last term is ${a_{22}}$ So, the ${n^{th}}$ term of an A.P is given by
${a_n} = a + (n - 1)d$
On putting all the values we have
$x = a + (22 - 1) \times 3$
On further solving, we get

$\therefore x = 64$

Note:
Arithmetic Progression in short AP is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. In mathematics, there are three types of progressions that are: Arithmetic Progression (AP), Harmonic Progression (HP) and Geometric Progression (GP).