Question

the sum of three consecutive terms in GP is 91. if 1,2 and 4 are subtracted from these numbers respectively, then the resulting numbers form an AP. find those three numbers which are in GP

Answer 1

Answer:

hence no. are 3,7,11 & 12,7,2.

Step-by-step explanation:

let three no. AP are

a-d , a , a+d

then their sum = a-d +a+ a+d = 21

3a =21

a=7

now acc to ques . no. becomes

a-d , a-1 , a+d+1

i.e 7-d, 6 , 8 +d are in gp

therefore 6/ 7-d = 8+d/6

36= (7-d)(8+d)

36= 56 + 7d -8d-d2

d2 +d -20=0 ---> d2 + 5d - 4d- 20=0 --->

d(d+5 ) -4(d+5)=0

d=4,-5

hence no. are 3,7,11 & 12,7,2.