_4+_1+2........ . .+x=437 solve the equation
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Sia ? 4 years, 9 months ago
(-4) + (-1) + 2 + 5 + ---- + x = 437.
Now,
-1 - (-4) = -1 + 4 = 3
2 - (-1) = 2 + 1 = 3
5 - 2 = 3
Thus, this forms an A.P. with a = -4, d = 3,l = x
Let their be n terms in this A.P.
Then,
Sn = {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} 874 = n[-8 + 3n - 3]
{tex}\Rightarrow{/tex}874 = n[3n - 11]
{tex}\Rightarrow{/tex}874 = 3n2 - 11n
{tex}\Rightarrow{/tex}3n2 - 11n - 874 = 0
{tex}\Rightarrow{/tex}3n2 - 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} or n = 19
Numbers of terms cannot be negative or fraction.
{tex}\Rightarrow{/tex} n = 19
Now, Sn = {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}
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