Question

Solve the equation: $1+4+7+10+....+x=287$ and find the value of x .

Answer 1

Hint: In this question, we can find that the series is in arithmetic progression. So, by using the formula for the sum of n terms in an arithmetic progression we get and equation which on solving gives the value of x.
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=\dfrac{n}{2}\left[ a+l \right]$

Complete step-by-step answer:

ARITHMETIC PROGRESSION: A sequence in which the difference between two consecutive terms is constant, is called an arithmetic progression (AP).
Sum of n terms of an AP is given by the formula
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=\dfrac{n}{2}\left[ a+l \right]$
Where, n is the number of terms
a is the first term of the sequence
d is the common difference of the sequence
The given sequence in the question is in AP as the difference between the consecutive terms is constant.
Now, from the given sequence in the question we have

& {{S}_{n}}=287 \\\ & a=1 \\\ & d=3 \\\ \end{aligned}$$ Now, by substituting these values in the above sum of n terms formula we get, $$\Rightarrow 287=\dfrac{n}{2}\left[ 2\times 1+\left( n-1 \right)\times 3 \right]$$ Now, by cross multiplying and on further simplification we get, $$\begin{aligned} & \Rightarrow 287\times 2=n\left[ 2+3n-3 \right] \\\ & \Rightarrow 574=3{{n}^{2}}-n \\\ \end{aligned}$$ Now, on rearranging the terms in the above quadratic equation we get, $$\Rightarrow 3{{n}^{2}}-n-574=0$$ Now, by solving the above equation using the direct formula we get, $$\Rightarrow n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ Now, on substituting the respective values we get, $$\Rightarrow n=\dfrac{1\pm \sqrt{1-4\times 3\times \left( -574 \right)}}{2\times 3}$$ Now, on further simplification we get, $$\Rightarrow n=\dfrac{1\pm \sqrt{1+6888}}{6}$$ $$\Rightarrow n=\dfrac{1\pm 83}{6}$$ $$\begin{aligned} & \Rightarrow n=\dfrac{1+83}{6},\dfrac{1-83}{6} \\\ & \Rightarrow n=14,\dfrac{-41}{3} \\\ \end{aligned}$$ Now, we can consider only the value of n which is integer because the number of terms in any sequence cannot be fractional. $$\therefore n=14$$ Now, in the given question x is the last term of the series. So, by using the formula for the sum of n terms of an AP in terms of the last term we can get the value of x. $$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$$ Here, $$\begin{aligned} & n=14 \\\ & {{S}_{n}}=287 \\\ & a=1 \\\ & l=x \\\ \end{aligned}$$ Now, by substituting the respective values in the above equation we get, $$\Rightarrow 287=\dfrac{14}{2}\left[ 1+x \right]$$ Now, this can be further written as $$\Rightarrow 287=7\left[ 1+x \right]$$ Let us now divide with 7 on both sides. $$\Rightarrow \dfrac{287}{7}=1+x$$ $$\begin{aligned} & \Rightarrow 41=1+x \\\ & \therefore x=40 \\\ \end{aligned}$$ Note: Instead of using the sum of n terms of an AP in terms of the last term formula to get the value of x we can directly get that from the last term of an AP series formula which is the same as the nth term of an AP series as we know the value of n. Both the methods give the same result. It is important to note that while finding the value of n from the sum of n terms of an AP formula we should not neglect any of the terms while rearranging or making operational mistakes because it changes the value of n and so the value of x also changes accordingly.