Question

Solve: $1+6+11+16+....+x=148$

Answer 1

The given arithmetic sequence is $1+6+11+16+....+x$ in which the first term is $a_1=1$, second term is $a_2=6$, $n$th term is $T_n=x$ and the sum of the series is $S_n=148$.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=6-1=5$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, with $a=1,d=5$ and $T_n=x$, we have

$T_n=a+(n-1)d\Rightarrow x=1+(n-1)5\Rightarrow x=1+5n-5\Rightarrow x=5n-4\Rightarrow 5n=x+4\Rightarrow n=\frac{x+4}{5}$

We also know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now to find the sum of series, substitute $n=\frac{x+4}{5},a=1,d=5$ and $S_n=148$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 148=\frac{\frac{x+4}{5}}{2}\left[\right(2\times 1)+(\frac{x+4}{5}-1)5]\Rightarrow 148=\frac{x+4}{5\times 2}[2+\left(\frac{x+4-5}{5}\right)5]\Rightarrow 148=\frac{x+4}{10}[2+(\frac{x-1}{5}\times 5)]$

$\Rightarrow 148\times 10=(x+4)(2+x-1)\Rightarrow 1480=(x+4)(x+1)\Rightarrow x^2+x+4x+4=1480\Rightarrow x^2+5x+4-1480=0\Rightarrow x^2+5x-1476=0$

$\Rightarrow x^2-36x+41x-1476=0\Rightarrow x(x-36)+41(x-36)=0\Rightarrow (x+41)(x-36)=0\Rightarrow x=-41,x=36$

Ignoring the negative value of $x$, we get $x=36$.

Hence, the last term of the series is $36$.