Which term of the $AP: -2, -7, -12, \ldots$ will be $-77$? Find the sum of this $AP$ up to the term $-77$.
(D) Given $AP$ is $-2, -7, -12, \ldots$
Let the $n^{th}$ term of the $AP$ be $T_n = -77$.
The first term $a = -2$ and the common difference $d = -7 - (-2) = -5$.
The formula for the $n^{th}$ term is $T_n = a + (n - 1)d$.
Substituting the values: $-77 = -2 + (n - 1)(-5)$.
$-77 + 2 = (n - 1)(-5) \Rightarrow -75 = (n - 1)(-5)$.
$n - 1 = \frac{-75}{-5} = 15 \Rightarrow n = 16$.
So,the $16^{th}$ term of the $AP$ is $-77$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}[a + l]$,where $l$ is the last term.
$S_{16} = \frac{16}{2}[-2 + (-77)] = 8[-79] = -632$.
Thus,the sum of this $AP$ up to the term $-77$ is $-632$.
Let the $n^{th}$ term of the $AP$ be $T_n = -77$.
The first term $a = -2$ and the common difference $d = -7 - (-2) = -5$.
The formula for the $n^{th}$ term is $T_n = a + (n - 1)d$.
Substituting the values: $-77 = -2 + (n - 1)(-5)$.
$-77 + 2 = (n - 1)(-5) \Rightarrow -75 = (n - 1)(-5)$.
$n - 1 = \frac{-75}{-5} = 15 \Rightarrow n = 16$.
So,the $16^{th}$ term of the $AP$ is $-77$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}[a + l]$,where $l$ is the last term.
$S_{16} = \frac{16}{2}[-2 + (-77)] = 8[-79] = -632$.
Thus,the sum of this $AP$ up to the term $-77$ is $-632$.
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