– 2 , - 7, - 12, ……
Let the nth term of the given AP be – 77.
First term, a = - 2 and common difference, d = - 7 – ( - 2 ) = - 7 + 2 = - 5
nth term of an AP, Tn = a + ( n – 1) d
⇒ −77=−2+(n−1)(−5)
⇒ −75=−(n−1)×5
⇒ (n−1)=15⇒n=16
So, the 16th term of the given AP will be -77.
Sum of n terms of an AP;
Sn=n2[2a+(n−1)d]
So, sum of 16 terms i.e., upto the term – 77;
i.e., S16=162[2×(−2)+(n−1)(−5)]
=8[−4+(16−1)(−5)]=8(−4−75)
=8×−79=−632
Hence, the sum of this AP upto the term – 77 is – 632.