Question

In G.P which begins with 1, the sum of three terms is 13, find the common ratio.

Answer 1

as per the question first three terms in gp are
$1 \: r \: {r}^{2}$
where r is the common ratio

now
$1 + r + {r}^{2} = 13$
solving
(r+4)(r-3)=0
we get
r=3 or r=-4

Answer 2

Answer:

a=1

now using the formula of sum of g .p = a/r+a+ar

now pitting the value of a =1

1 /r+1+1r =13

1+1r+1r^2=13

1 +1r+1r^2-13=0

1r^2+1r-12=0

1r^2+4r-3r-12=0

1r (r+4)-3(r+4)=0

(1r-3)(r+4)

therefore, r= 3 or r= -4 ANSWER

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