Question

How do you

find sum of
'n terms of an Arthimetic series.
Derive on
find out the
formula
​

Answer 1

Answer:

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] ← Eq. (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] ← Eq. (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

Answer Suppose, a1, a2, a3, ……….. be an Arithmetic Progression whose first term is a and common difference is d. Therefore, S = n2[2a + (n - 1)d] = n2[a {a + (n - 1)d}] = n2[a + l]. We can also find find the sum of first n terms of an Arithmetic Progression according to the process below.