Question
Find the sum:
$1 + (-2) + (-5) + (-8) + ........ + (-236)$
$1 + (-2) + (-5) + (-8) + ........ + (-236)$
✓
Answer
Here, first term (a) = 1 and common difference (d) = (-2) - 1 = -3 Sum of n terms of an
AP, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+\big(\text{n}-1\big)\text{d}\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\times1+\big(\text{n}-1\big)\times\big(-3\big)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big(2-3\text{n}+3\big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big(5-3\text{n}\big)........(\text{i})$
we know that, if the last term (l) of an AP is known, then l = a + (n - 1)
[$\because$ l = -236, given]
$\Rightarrow -236 = 1 + (n - 1)(-3)$
$\Rightarrow -237 = -(n - 1) \times 3$
$\Rightarrow n - 1 = 79$
$\Rightarrow n = 80$
Now, put the value of n in Eq. (i), we get
$\Rightarrow\text{S}_\text{n}=\frac{\text{80}}{2}\big[5-3\times80\big]$
$\Rightarrow S_n = 40(5 - 240)$
$\Rightarrow S_n = 40 \times (-235)$
$\Rightarrow S_n = -9400$
Hence, the required sum is -9400.Alternate method:
Given, a = 1, d = -3 and l = -236 Sum of n terms of an
AP, $\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$ [ $\because$ n = 80] $\Rightarrow\text{S}_\text{n}=\frac{\text{80}}{2}\big[1+\big(-236\big)\big]$
$\Rightarrow\text{S}_\text{n}=40\times(-235)$
$\Rightarrow\text{S}_\text{n}=-9400$
AP, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+\big(\text{n}-1\big)\text{d}\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\times1+\big(\text{n}-1\big)\times\big(-3\big)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big(2-3\text{n}+3\big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big(5-3\text{n}\big)........(\text{i})$
we know that, if the last term (l) of an AP is known, then l = a + (n - 1)
[$\because$ l = -236, given]
$\Rightarrow -236 = 1 + (n - 1)(-3)$
$\Rightarrow -237 = -(n - 1) \times 3$
$\Rightarrow n - 1 = 79$
$\Rightarrow n = 80$
Now, put the value of n in Eq. (i), we get
$\Rightarrow\text{S}_\text{n}=\frac{\text{80}}{2}\big[5-3\times80\big]$
$\Rightarrow S_n = 40(5 - 240)$
$\Rightarrow S_n = 40 \times (-235)$
$\Rightarrow S_n = -9400$
Hence, the required sum is -9400.Alternate method:
Given, a = 1, d = -3 and l = -236 Sum of n terms of an
AP, $\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$ [ $\because$ n = 80] $\Rightarrow\text{S}_\text{n}=\frac{\text{80}}{2}\big[1+\big(-236\big)\big]$
$\Rightarrow\text{S}_\text{n}=40\times(-235)$
$\Rightarrow\text{S}_\text{n}=-9400$
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