Consider three distinct real numbers a,b,c in a G.P with `a^2+b^2+c^2=t^2` and a+b+c =`alpha t` .The sum of the common ratio and its reciprocal is denoted by S.
Complete set of `alpha^2` is

Options

A. `alpha: beta = 2:1`
B. If `alpha " and " beta ` are roots of the equation `ax^2+bx+c =0` then `2 b^2 = 9ac`
C. The sum of infinite `G.P 1+r+r^2 + …. Is 3//2`
D. If `alpha =1` , then sum of 10 terms of A.P is 100

Answer

B, C, D

Explanation

Let `S_(n)` denote the sum of first n terms of A.P..
Accoording to the question ,
`r=(S_(n))/(S_(2n)-S_(n))=1/((S_(2n))/(S_(n))-1)` is independent of n.
Therefore, `(S_(2n))/(S_(n))` is independent of n.
Now, `(S_(2n))/(S_(n))=((2n)/2(2alpha+(2n-1)beta))/(n/2(2alpha+(n-1)beta))=(2(2alpha-beta+2nbeta))/((2alpha-beta+nbeta))`
The ratio of independent of n if
`2alpha-beta=0`
`thereforealpha : beta= 1 : 2`
`therefore r=1/(4-1)=1/3`
`alpha and beta` are the roots of the equation `ax^(2)+bx+c=0`
`therefore2alpha+alpha=-b/a and 2alphaxxalpha=c/a`
Solving these, we get
`2(-b/(3a))^(2)=c/a`
`rArr2b^(2)=9ac`
`1+r+r^(2)+...=1/(1-r)=1/(1-1/3)=3/2`
`alpha=1 and beta=2`
`therefore` Sum of 10 terms A.P.=`10/2(2xx1+(10-1)xx2)=100.`