Can you find x?

The series of numbers is an arithmetic progression . The first step is to find the number of terms, $n$ , from the formula

$S=\dfrac{n}{2}[2a_1+(n-1)d]$ $\large \implies$ $1344=\dfrac{n}{2}[2(4)+(n-1)(6)]$ $\large \implies$ $2688=n(8+6n-6)$ $\large \implies$ $2688=2n+6n^2$

Now use quadratic formula to solve for $n$ , we get, $n=21$ .

Now solve for $x$ using the formula

$S=\dfrac{n}{2}(a_1+a_n)$ $\large \implies$ $1344=\dfrac{21}{2}(4+x)$ $\large \implies$ $\dfrac{2688}{21}=4+x$ $\large \implies$ $\color{#D61F06}\large \boxed{x=124}$

suppose, a=4 and d=6

so, $[ \frac {n}{2} ( 2 \times a ) + ( n-1 ) d]$ = $1344$

or, $3n^2 + n$ = $1344$ ...................(calculator)

or, $n =21$ .....................(negative value is not applicable)

so, 21st sequence is = $a+ (n-1)d$

                              = 124

$\frac { n }{ 2 } (2a+(n-1)6)=1344\\ n(4+(n-1)3)=1344\\ n(4+3n-3)=1344\\ 3{ n }^{ 2 }+n=1344\\ 3{ n }^{ 2 }+n-1344=0$

${ n }_{ 1 },{ n }_{ 2 }=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ { n }_{ 1 },{ n }_{ 2 }=\frac { -1\pm \sqrt { { 1 }^{ 2 }-4\cdot 3\cdot (-1344) } }{ 6 } \\ { n }_{ 1 },{ n }_{ 2 }=\frac { -1\pm \sqrt { 1+5376 } }{ 6 } \\ { n }_{ 1 },{ n }_{ 2 }=\frac { -1\pm 127 }{ 6 } \\ { n }_{ 1 }=\frac { -1+127 }{ 6 } =\frac { 126 }{ 6 } =21\\ { n }_{ 2 }=\frac { -1-127 }{ 6 } =-\frac { 128 }{ 6 } =-21\frac { 1 }{ 3 }$

So the value of $n$ is $21$ .

${ U }_{ 21 }=a+(21-1)b\\ { U }_{ 21 }=4+20\cdot 6\\ { U }_{ 21 }=124$ .