Question
Solve the question: $(-4) + (-1) + 2 + 5 + ..... + x = 437.$
✓
Answer
Suppose $x$ is $n^{th}$ term of the given $A.P.$
$a_n= x$
Here, $a = -4, d = 3.$
It is given that, $S_n= 437.$
$\Rightarrow\ \frac{\text{n}}{2}[2(-4)+(\text{n}-1)3]=437$
$\Rightarrow\ 3\text{n}^2-11\text{n}-874=0$
$\Rightarrow\ 3\text{n}^2-57\text{n}+46\text{n}-874=0$
$\Rightarrow\ 3\text{n}(\text{n}-19)+46(\text{n}-19)=0$
$\Rightarrow\ \text{n}=-\frac{46}{3}, 19$
Since, $n$ cannot be in fraction so $n = 19.$
Now $a_n= x$
$⇒ (-4) + (19 - 1)3 = x$
$⇒ -4 + 54 = x$
$⇒ x = 50.$
$a_n= x$
Here, $a = -4, d = 3.$
It is given that, $S_n= 437.$
$\Rightarrow\ \frac{\text{n}}{2}[2(-4)+(\text{n}-1)3]=437$
$\Rightarrow\ 3\text{n}^2-11\text{n}-874=0$
$\Rightarrow\ 3\text{n}^2-57\text{n}+46\text{n}-874=0$
$\Rightarrow\ 3\text{n}(\text{n}-19)+46(\text{n}-19)=0$
$\Rightarrow\ \text{n}=-\frac{46}{3}, 19$
Since, $n$ cannot be in fraction so $n = 19.$
Now $a_n= x$
$⇒ (-4) + (19 - 1)3 = x$
$⇒ -4 + 54 = x$
$⇒ x = 50.$
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