Which term of the AP – 2, – 7, – 12,…

Given, AP -2,-7,-12,…

Let the nth term of an AP is – 77.

Then, first term (a) = – 2 and common difference (d) = – 7 – (- 2) = – 7 + 2 = – 5.

∴nth term of an AP, Tn = a + (n – 1)d

$\Rightarrow \quad-77=-2+(n-1)(-5)$

$\Rightarrow \quad-75=-(n-1) \times 5$

$\Rightarrow \quad(n-1)=15 \Rightarrow n=16$

So, the 16th term of the given AP will be $-77$.

Now the sum of $n$ terms of an $A P$ is

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

So, sum of 16 terms i.e., upto the term $-77$.

i.e.                                                  $S_{16}=\frac{16}{2}[2 \times(-2)+(n-1)(-5)]$

$=8[-4+(16-1)(-5)]=8(-4-75)$

$=8 \times-79=-632$

Hence, the sum of this AP upto the term – 77 is – 632.