1.n th term =a+(n-1)d (arithmetic)
2.Σ of n th term =n/2{2a+(n-1)d}
3.Σ of n th term =n/2(a+p)
4.n=(p-a)/d+1
5.1+2+3+...+n=n(n+1)/2
6.1²+2²+3²+...+n²=n{(n+1)(2n+1]}/6
7.1³+2³+3³+...n³=[n(n+1)/2]²
8.1+3+5+...+n=n²
9.2+4+6+...+n=n(n+1)
p=last term
a=first term
n=number of term
→2b=a+c
→b=(a+c)/2
∞=infinity
Σ=summation
r=common ratio
1.n th term=arⁿ⁻¹
2.Σ of n th term=a(rⁿ -1)/r-1 [r>1]
3.Σ of n th term=a(1-rⁿ )/1-r [r<1]
4.Σ of n th term=a/1-r [r=∞]
5.Σ of n th term=an [r=1]
6.a+b+c+d+... Then,a/b=b/c
→b²=ac