Question

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→If S1,S2 and S3 be the respectively the sum of n,2n and 3n terms of G.P. Prove that S1(S3-S2)=(S2-S1)².

-_-" help!!​

Answer 1

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S1 ,S2 and S3 are respectively the sum of n ,2n and 3n terms of a G.P

Lett say first term = a & common Ratio = r

S1 = a(rⁿ - 1)/(r-1)

S2 = a(r²ⁿ - 1)/(r-1)

S3 = a(r³ⁿ - 1)/(r-1)

LHS = S1(S3 - S2)

= {a(rⁿ - 1)/(r-1)} (a(r³ⁿ - 1)/(r-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)²r²ⁿ/ (r-1)²

= ( arⁿ(rⁿ - 1)/(r-1) )²

RHS = (S2-S1)²

= (a(r²ⁿ - 1)/(r-1) - a(rⁿ - 1)/(r-1))²

= (a/(r-1))²(r²ⁿ - 1 - rⁿ + 1)²

= (a/(r-1))²(r²ⁿ - rⁿ)²

= (arⁿ/(r-1))²(rⁿ - 1)²

= (arⁿ(rⁿ - 1)/(r-1))²

LHS = RHS

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Answer 2

$\huge{\underline{\underline{\mathbb{\color{grey}{SOLUTION:-}}}}}$

S1 ,S2 and S3 are respectively the sum of n ,2n and 3n terms of a G.P

Lett say first term = a & common Ratio = r

S1 = a(rⁿ - 1)/(r-1)

S2 = a(r²ⁿ - 1)/(r-1)

S3 = a(r³ⁿ - 1)/(r-1)

LHS = S1(S3 - S2)

= {a(rⁿ - 1)/(r-1)} (a(r³ⁿ - 1)/(r-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)²r²ⁿ/ (r-1)²

= ( arⁿ(rⁿ - 1)/(r-1) )²

RHS = (S2-S1)²

= (a(r²ⁿ - 1)/(r-1) - a(rⁿ - 1)/(r-1))²

= (a/(r-1))²(r²ⁿ - 1 - rⁿ + 1)²

= (a/(r-1))²(r²ⁿ - rⁿ)²

= (arⁿ/(r-1))²(rⁿ - 1)²

= (arⁿ(rⁿ - 1)/(r-1))²

LHS = RHS

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