Question

If S₁, S₂, and S₃ represent the number of n terms, 2n terms, and 3n of the first term of the geometry series, prove that:
A. S₁ (S₃ - S₂) = (S₂ - S₁) ²
B. S₁² + S₂² = S₁ (S₂ + S₃)​

Answer 1

Let a is the first term and r is the common ratio of GP .also S1 , S2 and S3 are the number of n terms , 2n terms and 3n terms of the first terms of the GP .

so,

S1 = a.(rⁿ -1)/(r -1)

S2 = a.(r²ⁿ-1)/(r -1)

S3 = a.(r³ⁿ-1)/(r -1)

A) LHS = S1(S3 - S2)

= a.(rⁿ-1)/(r-1) { a.(r³ⁿ-1)-a.(r²ⁿ-1)}/(r -1)

=a²(rⁿ-1)/(r-1)² { r³ⁿ -1 - r²ⁿ+1 }

= a².(rⁿ-1)/(r-1)².r²ⁿ(rⁿ-1)

= a².(rⁿ-1)².r²ⁿ/(r-1)²

=a².(rⁿ.rⁿ -1.rⁿ)²/(r-1)²

= a².(r²ⁿ-rⁿ)²/(r -1)²

= a²{(r²ⁿ-1) -(rⁿ-1)}/(r -1)²

= { a.(r²ⁿ-1)/(r -1) - a.(rⁿ-1)/(r -1)}²

= {S2 - S1 }² = RHS

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B) LHS = S1² + S2²

= {a.(rⁿ-1)/(r-1)}² + {a.(r²ⁿ-1)/(r -1)}²

= a²/(r-1)² { (rⁿ-1)² + (r²ⁿ-1)²}

=a²/(r -1)²{ r²ⁿ+1 -2rⁿ + r⁴ⁿ +1 -2r²ⁿ}

= a²/(r -1)² { 2 - 2rⁿ + r⁴ⁿ -r²ⁿ}

= a²/(r -1)²{ r²ⁿ( rⁿ -1)(rⁿ+1) -2(rⁿ -1) }

= a².(rⁿ -1)/(r -1)²{ r³ⁿ + r²ⁿ -2 }

= {a².(rⁿ-1)/(r -1)} { (r³ⁿ -1)/(r -1) + (r²ⁿ-1)/(r -1)}

= { a.(rⁿ-1)/(r -1)} { a.(r³ⁿ-1)/(r -1) + a.(r²ⁿ -1)/(r -1)}

= S1 ( S3 + S2) = RHS

Answer 2

Answer:

If S₁, S₂, and S₃ represent the number of n terms, 2n terms, and 3n of the first term of the geometry series, prove that:

A. S₁ (S₃ - S₂) = (S₂ - S₁) ²

B. S₁² + S₂² = S₁ (S₂ + S₃)