Question

Class 10
Ch - Arithmetic Progression
$\bf{Question-}$
Solve the equation:
$\sf{-4+(-1)+2+...+x=437}$
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Answer 1

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Answer 2

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Given equation : -4 + (-1) + 2 ..…. + x = 437

Here, a = -4

d = -1 - (-4)

=> d = 3

$\sf{S_n}$ = 437

We know that,

$\large{ \boxed{ \green{ \sf{S_n\:=\:\dfrac{n}{2}[2a + (n - 1)d]}}}}$

$\implies$ 437 = $\sf{ \dfrac{n}{2} [2(-4) + (n - 1)(3)]}$

$\implies$ 437 × 2 = n ( -8 + 3n - 3 )

$\implies$ 874 = -11n + 3n²

$\implies$ 3n² - 11n - 874 = 0

By splitting middle term,

$\implies$ 3n² - 57n + 46n - 874 = 0

$\implies$ 3n ( n - 19 ) + 46 ( n - 19 ) = 0

$\implies$ ( 3n + 46 ) ( n - 19 ) = 0

$\implies$ n = 19 and - $\dfrac{46}{3}$

Rejecting the negative values since number of terms can't be negative,

$\implies$ n = 19

Now,

x = 2a + ( n - 1 ) d

$\implies$ x = (-4) + ( 19 - 1 ) ( 3 )

$\implies$ x = -4 + 18 × 3

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