Question -
Answer -
Given:
Sum of G.P of 3 termsis 125
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
125 = a (rn –1)/(r-1)
125 = a (r3 –1)/ (r-1) … equation (1)
Now,
Sum of G.P of 6 termsis 152
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
152 = a (rn –1)/(r-1)
152 = a (r6 –1)/ (r-1) … equation (2)
Let us divide equation(i) by (ii) we get,
125/152 = [a (r3 –1)/ (r-1)] / [a (r6 – 1)/ (r-1)]
125/152 = (r3 –1)/(r6 – 1)
125/152 = (r3 –1)/[(r3 – 1) (r3 + 1)]
125/152 = 1/(r3 +1)
125(r3 +1) = 152
125r3 +125 = 152
125r3 =152 – 125
125r3 =27
r3 =27/125
r3 = 33/53
r = 3/5
∴ The common ratiois 3/5