if the sum of the first two terms of a go is 3 and the sum of the second and the third terms is-6 find the sum of the first term and the common ratio

Sum of first and third term is = -3 + -12 = 15

sum of all the three terms is = -3 + 6 + -12 = -9

common difference = -2

Explanation:

For the GP let the first term be a

and common ratio be r

n the term of gp is given by ar^(n-1)

then

first term = a

second term = ar^(2-1) = ar

third term = ar^(3-1)  = ar^2

given, sum of the first two terms of a gp is 3

a + ar = 3

=> a(1+r) = 3 > 1

also ,  sum of the second and the third terms is-6

ar + ar^2 = -6

taking ar as common

=>ar(1+r) = -6

it can be also written as product of r and a(1+r)

=> r*a(1+r)= -6

we have calculated value of a(1+r) as 3, substituting this value here we have

=> r*3 = -6

=> r = -6/3 = -2

Thus, common difference is -2.

As a(1+r) = 3  ,  substituting  value of r as -2  here, we have

a(1+(-2) = 3

=> a(-1)= 3

=> a = -3

Thus, first term is a = -3

second term is ar = -3*-2= 6

third term is ar^2 = -3*(-2)^2= -3*4 = -12

Sum of first and third term is = -3 + -12 = 15

sum of all the three terms is = -3 + 6 + -12 = -9

common difference = -2

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