UPSC CDS Math Trigonometric Ratios and Trigonometric Identities Questions

Sn=861
⇒‌‌

n(n+1)

2

=861
⇒n2+n−1722=0
⇒n2+42n−41n−1722=0
⇒n(n+42)−41(n+42)=0
⇒(n+42)(n−41)=0
⇒n=−42,41
Thus, there are exactly two values of n for which Sn=861.
Hence, statement 1 is correct.
Sn=S−(n+1)
⇒n=n
This is true for all values of n.
For m=–1,

n(n+1)

2

=−1⇒n2+n+2=0
But, there exists no real integral value of n which satisfies this equation.
Hence, statement 2 is incorrect.