Question
Solve the quation:
$-4 + (-1) + 2 + ...... + x = 437$
$-4 + (-1) + 2 + ...... + x = 437$
✓
Answer
Given list of numbers are in AP.
$[d = d_1 = d_2 = 3]$
$a = -4$ and $a_n= x$
$As a_n = x$
$\Rightarrow a + (n - 1)d = x$
$\Rightarrow -4 + (n - 1)(3) = x + 4$
$\Rightarrow\big(\text{n}-1\big)=\frac{\text{x}+4}{3}$
$\Rightarrow\text{n}=\frac{\text{x}+4}{3}+1$
$\Rightarrow\text{n}=\frac{\text{x}+4+3}{3}$
$\Rightarrow\text{n}=\frac{\text{x}+7}{3}......(\text{i})$
$\Rightarrow\text{S}_{\text{n}}=\frac{\text{n}}{2}\big[2\text{a}+\big(\text{n}-1\big)\text{d}\big]$
$\therefore\text{S}_{\text{n}}=\frac{\big(\text{x}+7\big)}{\big(2\times3\big)}\bigg[2\big(-4\big)+\frac{\big(\text{x}+4\big)3}{3}\bigg]\big[\text{Using(i)}\big]$
$\text{S}_{\text{n}}=\frac{\big(\text{x}+7\big)}{6}\big[-8+\text{x}+4\big]$
$\text{S}_{\text{n}}=\frac{\big(\text{x}+7\big)\big(\text{x}-4\big )}{6}$
$\text{S}_{\text{n}}=437$
$\Rightarrow x^2 + 3x - 28 - 2622 = 0$
$\Rightarrow x^2 + 3x - 2650 = 0$
$\Rightarrow x^2 + 53x - 50x - 2650 = 0$
$\Rightarrow x(x + 53) - 50(x + 53) = 0$
$\Rightarrow (x + 53)(x - 50) = 0$
$\Rightarrow x = -53 or x = 50$
Rejecting the negative value x = -53, we have x = -50.
So, x = 50 is the requried value as forward terms are positive.
$[d = d_1 = d_2 = 3]$
$a = -4$ and $a_n= x$
$As a_n = x$
$\Rightarrow a + (n - 1)d = x$
$\Rightarrow -4 + (n - 1)(3) = x + 4$
$\Rightarrow\big(\text{n}-1\big)=\frac{\text{x}+4}{3}$
$\Rightarrow\text{n}=\frac{\text{x}+4}{3}+1$
$\Rightarrow\text{n}=\frac{\text{x}+4+3}{3}$
$\Rightarrow\text{n}=\frac{\text{x}+7}{3}......(\text{i})$
$\Rightarrow\text{S}_{\text{n}}=\frac{\text{n}}{2}\big[2\text{a}+\big(\text{n}-1\big)\text{d}\big]$
$\therefore\text{S}_{\text{n}}=\frac{\big(\text{x}+7\big)}{\big(2\times3\big)}\bigg[2\big(-4\big)+\frac{\big(\text{x}+4\big)3}{3}\bigg]\big[\text{Using(i)}\big]$
$\text{S}_{\text{n}}=\frac{\big(\text{x}+7\big)}{6}\big[-8+\text{x}+4\big]$
$\text{S}_{\text{n}}=\frac{\big(\text{x}+7\big)\big(\text{x}-4\big )}{6}$
$\text{S}_{\text{n}}=437$
$\Rightarrow x^2 + 3x - 28 - 2622 = 0$
$\Rightarrow x^2 + 3x - 2650 = 0$
$\Rightarrow x^2 + 53x - 50x - 2650 = 0$
$\Rightarrow x(x + 53) - 50(x + 53) = 0$
$\Rightarrow (x + 53)(x - 50) = 0$
$\Rightarrow x = -53 or x = 50$
Rejecting the negative value x = -53, we have x = -50.
So, x = 50 is the requried value as forward terms are positive.
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