1.

The sum of the first ten terms of the geometric progression is `S_1` and the sum of the next ten terms (11th term to 20th term) is `S_2` then the common ratio will be:

Answer» `S_1=(a(r^10-11))/(r-1)-(1)`
r-common ratio
a= first number
`S_2=(a(r^n-1))/(r-1)-(a(r^10-1))/(r-1)`
`S_2=(a(r^20-1))/(r-1)-S_1`
`S_1+S_2=(a(r^20-1))/(r-1)-(2)`
`(r-1)=(a(r^10-1))/S_1`
`S_1+S_2=(a(r^10-1)(r^10+1))/(a(r^10-1))*S_1`
`r^10+1=1+S_2/S_1`
`r=10sqrt(S_2/S_1`.


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