Question

Find four numbers the first three of which form a GP and the last three form an AP. The sum of the extreme terms is 14 and the sum of middle term is 12.
Can anyone have brain to solve .
Solve in notebook then add your answer

Answer 1

Dear Student,

◆ Answer -

Four numbers are 12, 8, 4 and 2

◆ Explanation -

Sum of extreme terms is 14. So if 1st no is x, 4th no will be 12-x.

Sum of middle terms is 12. So if 2nd no is y, 3rd no will be 12-y.

Given that, 1st 3 terms are in AP,

y - x = 12-y - y

x = 3y - 12

Also, last 3 terms are in GP,

(12-y) / y = (14-x) / (12-y)

(12-y)(12-y) = y (14-3y+12)

144 - 24y + y^2 = 14y - 3y^2 + 12y

4y^2 - 50y + 144 = 0

2y^2 - 25y + 72 = 0

Solving this quadratic eqn,

y = 8

Now, the numbers will be ,

x = 3×8-12 = 12

y = 8

12-y = 12-8 = 4

14-x = 14-12 = 2

Therefore, the four numbers are 12, 8, 4 and 2.

Thanks dear...