Question

DERIVATION of SUM OF nTH term of an A.P.!

CLASS- 10
CHAPTER: ARITHMETIC PROGRESSIONS (A.P.)

Answer 1

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]


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Answer 2

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