NDA Math Sequences and Series

Concept:
If a1,a2,...,an be an AP then Sn=a1+a2+...+an=‌

n

2

×(2a+(n−1)d) where a is the 1‌st ‌ term and d is the common difference.
If a1,a2,...,an be an AP then the general term is given by: an=a+(n−1)×d where a is the 1‌st ‌ term and d is the common difference.
Calculation:
Here, we have to find the value of 1−2+3−4+5− +101 ⇒1−2+3−4+5− +101=(1+3+......+101)−(2+4+.....+100)
As, we can see that (1,3,......,101) is an AP with a=1 and d=2. ⇒an=101=1+(n−1)×2 ⇒n=51
⇒S51=1+3+...+101=‌

51

2

×(2+50×2)=2601
Similarly, (2,4,...,100) is an AP with a=2 and d=2
⇒an=100=2+(n−1)×2
⇒n=50
⇒S50=2+4+...+100=‌

50

2

×(4+49×2)=2550
⇒1−2+3−4+5−...−+101=(1+3+......+101)−(2+4+.....+100)=2601−2550=51