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# Solve :- $1+4+7+10+.....+x=590$?

Last updated date: 23rd Jun 2024
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Hint: we can solve this using arithmetic progression concept. We can see that the given expression is arithmetic progression. So we can find the sum of n terms using the formula and solve the equation. Then we can get a number of terms in the expression using the formula we have AP.

So let’s see the formulas of arithmetic progression to solve the problem.
Sum of n terms is given by
${{s}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
To find the last term
$x=a+\left( n-1 \right)d$

Given
$1+4+7+10+.....+x=590$
We can say that
$1,4,7,10....$ are in AP because
The common difference is
$d={{a}_{2}}-{{a}_{1}}=4-1$
$\Rightarrow 3$ is the same.
And the first term is $1$.
So we also have a sum of n terms by which we can find the number of terms in the series.
Sum of n terms is $590$.
The formula of sum of n terms is
${{s}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
By substituting the values we have , we will get
$590=\dfrac{n}{2}\left( 2\times 1+\left( n-1 \right)3 \right)$
Now we have to simplify the above equation.
First we will remove the inner parenthesis by multiplying with 3. We will get
$\Rightarrow 590=\dfrac{n}{2}\left( 2+n\left( 3 \right)-1\left( 3 \right) \right)$
$\Rightarrow 590=\dfrac{n}{2}\left( 2+3n-3 \right)$
By simplifying it we get
$\Rightarrow 590=\dfrac{n}{2}\left( 3n-1 \right)$
Now we have to multiply with 2 on both sides of the equation.
$\Rightarrow 590\times 2=\dfrac{n}{2}\left( 3n-1 \right)\times 2$
By simplifying it we will get
$\Rightarrow 1180=n\left( 3n-1 \right)$
Now we have to remove parenthesis on RHS by multiplying with n.
We will get
$\Rightarrow 1180=3{{n}^{2}}-n$
Now we have to subtract 1180 on both sides of the equation.
$\Rightarrow 1180-1180=3{{n}^{2}}-n-1180$
We will get
$\Rightarrow 3{{n}^{2}}-n-1180=0$
Now we have a quadratic equation now we have to solve it to find the n.
We will solve it by factoring method.
So we have to find prime factors for 1180 and have to split accordingly which satisfies the conditions.
We will get -60 and 59 as factors.
Now we have to split them we will get
$\Rightarrow 3{{n}^{2}}-60n+59n-1180=0$
By taking common terms out we will get
$\Rightarrow 3n\left( n-20 \right)+59\left( n-20 \right)=0$
$\Rightarrow \left( n-20 \right)\left( 3n+59 \right)=0$
We will get n as
$n=20$
$n=\dfrac{-59}{3}$
N cannot be negative so the value of n is 20.
Using this n we can get the last term.
The formula is
$x=a+\left( n-1 \right)d$
By substituting the values we get
$\Rightarrow x=1+\left( 20-1 \right)3$
$\Rightarrow x=1+\left( 19 \right)3$
By simplifying we will get
$\Rightarrow x=1+57$
$\Rightarrow x=58$
So the last term is 58.

Note: We can solve the equation using quadratic formula also to find the value of n. we should be aware of formulas to find the solution. When these types of series are given we have to check whether it is AP or GP then we can solve the question easily.