Question

If Sn, S2n, S3n are the sum of n, 2n , 3n terms of a GP respectively, then verify that:
Sn(S3n-S2n) =(S2n-Sn) ²

Answer 1

Answer:

The answer is S1 (S3-S2)=-(S2-S1)²

Step-by-step explanation:

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

Let's say first term = a

And 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

Answer 2

Answer:

Sn (S3n-S2n)=-(S2n-Sn)²

Step-by-step explanation:

Sn ,S2n and S3n are respectively the sum of n ,2n and 3n terms of a G.P

Let say first term = a  & common Ratio = r

Sn = a(rⁿ - 1)/(r-1)

S2n = a(r²ⁿ - 1)/(r-1)

S3n = a(r³ⁿ - 1)/(r-1)

LHS = Sn(S3n - S2n)

= {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 = (S2n-Sn)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