Question

Solve the equation ( - 4 ) + ( - 1 ) + 2 + .... + x = 437.

Answer 1

$\underline{\mathfrak{Solution : }}$


$\textsf{The above series forms an A.P.}$



$\mathsf{Where, } \\ \\ \mathsf{ \implies First \: term(a) \: = \: -4 } \\ \\ \mathsf{\implies Last \: term (l) \: = \: x } \\ \\ \mathsf{\implies Common \: difference (d) \: = \: (-1) \: - \: (-4)} \\ \\ \mathsf{ \qquad \qquad \qquad \qquad \qquad \qquad \: = \: -1 \: + \: 4 }\: \\ \\ \mathsf{ \qquad \qquad \qquad \qquad \qquad \qquad \: = \: 3 }$


$\mathsf{Let, no. \: of \: terms \: is \: n. }$




$\mathsf{Using \: Formula, } \\ \\ \mathsf{ \implies l \: = \: a \: + \: (n \: - \: 1)d } \\ \\ \mathsf{\implies x \: = \: (-4) \: + \: (n \: - \: 1)3 } \\ \\ \mathsf{\implies x \: = \: -4 \: + \: 3n \: - \: 3 } \\ \\ \mathsf{\implies x\: = \: 3n \: - \: 7 } \\ \\ \mathsf{\implies x\: + \: 7 \: = \: 3n } \\ \\ \mathsf{\therefore \quad n \: = \: \dfrac{( x \: + \: 7)}{3}}$



$\underline{\mathsf{Now, }} \\ \\ \mathsf{\implies S_{n} \: = \: \dfrac{n}{2}( a \: + \: l )} \\ \\\textsf{Plug the value of \textbf{n}, } \\ \\ \\ \mathsf{\implies 437 \: = \: \dfrac{ \quad(\dfrac{x\: + \: 7}{3}) \quad}{2}( -4 \: + \: x )}$




$\\ \mathsf{\implies 437 \: \times \: 2 \: = \: \dfrac{(x \: + \: 7)}{3}(x \: - \: 4 )} \\ \\ \mathsf{\implies 437 \times \: 2 \: \times \: 3 \: = \: (x \: + \: 7 )( x \: - \: 4 )}$


$\\ \mathsf{\implies 2622 \: = \: {x}^{2} \: - \: 4x \: + \: 7x \: - \: 28 } \\ \\ \mathsf{ \implies {x}^{2} \: + \: 3x \: - \: 28 \: - \: 2622 \: = \: 0 } \\ \\ \mathsf{ \implies {x}^{2} \: + \: 3x \: - \: 2650 \: = \: 0}$



$\mathsf{Here, } \\ \\ \mathsf{\implies Coefficient \: of \: {x}^{2}(a) \: = \: 1 } \\ \\ \mathsf{\implies Coefficient \: of \: x (b ) \: = \: 3 } \\ \\ \mathsf{\implies Constant \: term(c) \: = \: -2650 }$



$\textsf{Using Quadratic Formula , } \\ \\ \mathsf{ \implies x \: = \: \dfrac{ - b \: \pm \: \sqrt{ \: {b}^{2} \: - \: 4ac \: } }{2a} }$


$\\ \mathsf{ \implies x \: = \: \dfrac{ - 3 \: \pm \: \sqrt{ \: ( {3}^{2} ) \: - \: 4 \: \times \: 1 \: \times \: ( - 2650)} }{2 \: \times \: 1} } \\ \\ \\ \mathsf{ \implies x \: = \: \dfrac{ - 3 \: \pm \: \sqrt{ \: 9 \: + \: 10600} }{2}}$




$\\ \mathsf{ \implies x \: = \: \dfrac{ - 3 \: \pm \: \sqrt{ \: 10609} }{2} } \\ \\ \\ \mathsf{ \implies x \: = \: \dfrac{ - 3 \: \pm \: 103}{2}}$



$\\ \mathsf{\implies x \: = \: \dfrac{-3 \: + \: 103}{2} \quad, \implies x \: = \: \dfrac{-3 \: - \: 103}{2}} \\ \\ \\ \mathsf{\implies x \: = \: \dfrac{100}{2} \quad, \implies x \: = \: \dfrac{-106}{2}} \\ \\ \\ \mathsf{\implies x \: = \: 50 \: or \: (-53 ) \: [ Not \: Possible ]}$



$\boxed{\underline{\mathsf{Therefore, x \: is \: equal \: to \: 50.}}}$

Answer 2

Given Equation is (-4) + (-1) + 2 + .... + x = 437.

Observe carefully, the given equation is an arithmetic equation i.e arithmetic series.

Here, First term a = -4.

Common difference d = -1 + 4 = 3.

Sn = 437.

We know that sum of first n terms is given by:

⇒ Sn = (n/2)[2a + (n - 1) * d]

⇒ 437 = (n/2)[-8 + (n - 1) * 3]

⇒ 437  = (n/2)[-8 + 3n - 3]

⇒ 437 = (n/2)[3n - 11]

⇒ 874 = n[3n - 11]

⇒ 874 = 3n^2 - 11n

⇒ 3n^2 - 11n - 874 = 0

⇒ 3n^2 - 57n + 46n - 874 = 0

⇒ 3n(n - 19) + 46(n - 19) = 0

⇒ (n - 19)(3n + 46) = 0

⇒ n = 19,n = -46/3[neglect -ve values]

⇒ n = 19

Now,

We know that Sum of n terms is also equal to the formula:

⇒ Sn = (n/2)[a + l]

⇒ 437 = (19/2)[-4 + l]

⇒ 874 = 19[-4 + l]

⇒ 46 = -4 + l

⇒ 46 + 4 = l

⇒ l = 50.

Therefore, the final answer is 50.

Hope this helps!