Question

Sum of first 3 terms is to the sum of first 6 terms as 125:152. Find common ratio of GP

Answer 1

Sum of nth term in GP (Sn)= a(rⁿ -1)/(r -1)
A/C to question,
S3/S6 = 125/152
{a(r³ - 1)/(r -1)}/{a(r^6 -1)/(r -1)} = 125/152
(r³ -1)/(r³ -1)(r³+1) = 125/152
1/(r³ + 1) = 125/152
152 = 125r³ + 125
27 =125r³
27/125 = r³
(3/5)³ = r³
r = 3/5

hence, common ratio = 3/5

Answer 2

Explanation:Sn=arn−1r−1inGP

Giventhat

S3S6=a(r3−1/r−1)a(r6−1/r−1)=125/152

Assumingr≠1

(asr−1isindenominatorξdenomitorcannotbe0,r−1≠0,r≠1)

a(r3−1/r−1)a(r6−1/r−1)=125/152

r3−1r6−1=125/152

152r3−152=125r6−125

125r6−152r3−27=0

125r3(r3−1)−27(r3−1=0)

(125r3−27)(r3−1)=0

125r3=27orr3−1=0

r3=27/125r=1

r=3/5r≠1

Answer=r=3/5iscommonratio.