Question

Solve the given equation to find the value of ‘x’.
1 + 4 + 7 + 10 + ……….. + x = 287

Answer 1

Hint: First find the type of sequence which is formed by the given sequence. Find the value of n such that the sum of n terms of series is a given number. Then find the ${{n}^{th}}$ term of that series by using the n value you got before. This term is the required solution for a given question.

Complete step-by-step answer:
The give sum in the question is written in form of,
1 + 4 + 7 + ………….. + x = 287
The sequence of terms in the left hand side are given by, 1, 4, 7…………….
The difference between the first 2 terms, can be written as 4 – 1 = 3. The difference between the next 2 terms, can be written as 7 – 4 = 3. As these both are equal we can say it is an arithmetic progression. A sequence of numbers such that the difference of any two consecutive numbers is constant is called an arithmetic progression. For example the sequence 1, 2 , 3…………. Is an arithmetic progression with common difference = 2 – 1 = 1. Now, we need to find the ${{n}^{th}}$ term of such sequence. Let us have a sequence with first term a and common difference d.
We know the difference between successive terms is d. So, second term = first term + d = a + d.
Third term = second term + d = a + 2d and so on calculating, we get
${{n}^{th}}={{\left( n-1 \right)}^{th}}\text{term}+d=a+\left( n-2 \right)d+d=a+\left( n-1 \right)d$
The sum of n terms in an arithmetic progression S, can be written as
$S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Let the x be ${{n}^{th}}$ term of sequence given. So, we can write as
$1+4+7+..............+x=\dfrac{n}{2}\left( 2a\left( n-1 \right)d \right)$
By substituting a as 1 and d as 3, we can write it as
$1+4+7+..............+x=\dfrac{n}{2}\left( 2\left( n-1 \right)3 \right)$
By equating above equation to 287, we can write it as
$\dfrac{n}{2}\left( 2+3n-3 \right)=287$
By simplifying the above equation, we can write it as follows
$\dfrac{n}{2}\left( 3n-1 \right)=287$
By multiplying with 2 on both sides, we get it in form
$n\left( 3n-1 \right)=287\times 2$
By multiplying n inside the bracket, we can write the equation as
$3{{n}^{2}}-n=287\times 2$
By simplifying above equation, we can write equation as
$3{{n}^{2}}-n-574=0$
By quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ , we get values of n as
$n=\dfrac{1\pm \sqrt{1-4\left( 3 \right)\left( -574 \right)}}{6}=\dfrac{1\pm \sqrt{6889}}{6}$
By simplifying, we can write values of n as given below
$n=14,\dfrac{-41}{3}$
As n cannot be a fraction, n must be 14. By the ${{n}^{th}}$ term formula, we can write value of x as
x = a + (14 – 1)d = 1 + (14 – 1)3
By simplifying, we can write value of x in form of
$x=1+13\times 3$
By simplifying we get the value of x in the form of x = 40. Therefore the value of x in the question is 40.

Note: The use of formula of arithmetic progression must be done carefully because values of a, d are important to calculate ${{n}^{th}}$ term, sum of n terms. The quadratic equation formula has “-“ inside the square root. Some students generally take it as + and get the wrong result. So, be careful at that point.