Thank you
Its not wrong, it is a question from APFirst term is -4 , xommon difference is 3,Sn is 437.First find the value of n by sum of n terms formula and then find nth term
I am sure its wrong ques

(-4) + (-1) + 2 + 5 + —- + x = 437.Now,-1 – (-4) = -1 + 4 = 32 – (-1) = 2 +\xa01 = 35 – 2 = 3Thus, this forms an A.P. with a = -4, d = 3,l = xLet their be n terms in this A.P.Then,Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ 2 a + ( n – 1 ) d ] {/tex}{tex}\\Rightarrow 437 = \\frac { n } { 2 } [ 2 \\times ( – 4 ) + ( n – 1 ) \\times 3 ]{/tex}{tex}\\Rightarrow{/tex}\xa0874 = n[-8 + 3n – 3]{tex}\\Rightarrow{/tex}874 = n[3n – 11]{tex}\\Rightarrow{/tex}874 = 3n2\xa0- 11n{tex}\\Rightarrow{/tex}3n2\xa0- 11n – 874 = 0{tex}\\Rightarrow{/tex}3n2\xa0- 57n + 46n – 874 = 0{tex}\\Rightarrow{/tex}3n(n – 19) + 46(n – 19) = 0{tex}\\Rightarrow{/tex}3n + 46 = 0 or n = 19{tex}\\Rightarrow n = – \\frac { 46 } { 3 }{/tex}\xa0or n\xa0= 19Numbers of terms cannot be negative or fraction.{tex}\\Rightarrow{/tex}\xa0n = 19Now, Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ a + l ]{/tex}{tex}\\Rightarrow 437 = \\frac { 19 } { 2 } [ – 4 + x ]{/tex}{tex}\\Rightarrow – 4 + x = \\frac { 437 \\times 2 } { 19 }{/tex}{tex}\\Rightarrow – 4 + x = 46{/tex}{tex}\\Rightarrow x = 50{/tex}

It is an arithmetic series.a = -4, d = +3-4 – 1 + 2 + 5 + …. +\xa0(x – 3) + x = 437Sum of the series = [\xa02 a + (n-1)d ]n /2 = 437\xa0[- 8 + 3(n-1)] n / 2 = 437- 11n + 3n² = 8743 n² -11n – 874 = 0n = (11 +- 103 )/6 = 114/6 = 19 as n is positive\xa0only.So, x = a + (n-1)d = -4 + 3 (19-1) = 50

(-4) + (-1) + 2 + 5 + —- + x = 437.Now,-1 – (-4) = -1 + 4 = 32 – (-1) = 2 +\xa01 = 35 – 2 = 3Thus, this forms an A.P. with a = -4, d = 3,l = xLet their be n terms in this A.P.Then,Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ 2 a + ( n – 1 ) d ] {/tex}{tex}\\Rightarrow 437 = \\frac { n } { 2 } [ 2 \\times ( – 4 ) + ( n – 1 ) \\times 3 ]{/tex}{tex}\\Rightarrow{/tex}\xa0874 = n[-8 + 3n – 3]{tex}\\Rightarrow{/tex}874 = n[3n – 11]{tex}\\Rightarrow{/tex}874 = 3n2\xa0- 11n{tex}\\Rightarrow{/tex}3n2\xa0- 11n – 874 = 0{tex}\\Rightarrow{/tex}3n2\xa0- 57n + 46n – 874 = 0{tex}\\Rightarrow{/tex}3n(n – 19) + 46(n – 19) = 0{tex}\\Rightarrow{/tex}3n + 46 = 0 or n = 19{tex}\\Rightarrow n = – \\frac { 46 } { 3 }{/tex}\xa0or n\xa0= 19Numbers of terms cannot be negative or fraction.{tex}\\Rightarrow{/tex}\xa0n = 19Now, Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ a + l ]{/tex}{tex}\\Rightarrow 437 = \\frac { 19 } { 2 } [ – 4 + x ]{/tex}{tex}\\Rightarrow – 4 + x = \\frac { 437 \\times 2 } { 19 }{/tex}{tex}\\Rightarrow – 4 + x = 46{/tex}{tex}\\Rightarrow x = 50{/tex}

(-4) + (-1) + 2 + 5 + —- + x = 437.Now,-1 – (-4) = -1 + 4 = 32 – (-1) = 2 +\xa01 = 35 – 2 = 3Thus, this forms an A.P. with a = -4, d = 3,l = xLet their be n terms in this A.P.Then,Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ 2 a + ( n – 1 ) d ] {/tex}{tex}\\Rightarrow 437 = \\frac { n } { 2 } [ 2 \\times ( – 4 ) + ( n – 1 ) \\times 3 ]{/tex}{tex}\\Rightarrow{/tex}\xa0874 = n[-8 + 3n – 3]{tex}\\Rightarrow{/tex}874 = n[3n – 11]{tex}\\Rightarrow{/tex}874 = 3n2\xa0- 11n{tex}\\Rightarrow{/tex}3n2\xa0- 11n – 874 = 0{tex}\\Rightarrow{/tex}3n2\xa0- 57n + 46n – 874 = 0{tex}\\Rightarrow{/tex}3n(n – 19) + 46(n – 19) = 0{tex}\\Rightarrow{/tex}3n + 46 = 0 or n = 19{tex}\\Rightarrow n = – \\frac { 46 } { 3 }{/tex}\xa0or n\xa0= 19Numbers of terms cannot be negative or fraction.{tex}\\Rightarrow{/tex}\xa0n = 19Now, Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ a + l ]{/tex}{tex}\\Rightarrow 437 = \\frac { 19 } { 2 } [ – 4 + x ]{/tex}{tex}\\Rightarrow – 4 + x = \\frac { 437 \\times 2 } { 19 }{/tex}{tex}\\Rightarrow – 4 + x = 46{/tex}{tex}\\Rightarrow x = 50{/tex}