Question

Derive formula for the nth term of an AP.

Answer 1

Derivation of Formulas
Let
dd = common difference
a1a1 = first term
a2a2 = second term
a3a3 = third term
...

amam = mth term or any term before anan
...

anan = nth term or last term
 

d=a2−a1=a3−a2=a4−a3d=a2−a1=a3−a2=a4−a3 and so on.
 

Derivation for an in terms of a1 and d
a1=a1a1=a1

a2=a1+da2=a1+d

a3=a2+d=(a1+d)+d=a1+2da3=a2+d=(a1+d)+d=a1+2d

a4=a3+d=(a1+2d)+d=a1+3da4=a3+d=(a1+2d)+d=a1+3d

a5=a4+d=(a1+3d)+d=a1+4da5=a4+d=(a1+3d)+d=a1+4d
...

am=a1+(m−1)dam=a1+(m−1)d
...

an=a1+(n−1)dan=a1+(n−1)d

 

In similar manner
an=anan=an

an−1=an−dan−1=an−d

an−2=an−1−d=(an−d)−d=an−2dan−2=an−1−d=(an−d)−d=an−2d

an−3=an−2−d=(an−2d)−d=an−3dan−3=an−2−d=(an−2d)−d=an−3d

an−4=an−3−d=(an−3d)−d=an−4dan−4=an−3−d=(an−3d)−d=an−4d
...

am=an−(n−m)dam=an−(n−m)d
...

a1=an−(n−1)da1=an−(n−1)d

 

Derivation for the Sum of Arithmetic Progression, S
S=a1+a2+a3+a4+...+anS=a1+a2+a3+a4+...+an

S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d]S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d]   →   Equation (1)
 

S=an+an−1+an−2+an−3+...+a1S=an+an−1+an−2+an−3+...+a1

S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d]S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d]   →   Equation (2)
 

Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)

2S=n(a1+an)2S=n(a1+an)

S=n2(a1+an)S=n2(a1+an)

 

Substitute an = a1 + (n - 1)d to the above equation, we have
S=n2{a1+[a1+(n−1)d]}S=n2{a1+[a1+(n−1)d]}

S=n2[2a1+(n−1)d]

Answer 2

Tn = a + (n-1)d
where
a = first term
d= common difference