Find the smallest number in a GP whose sum is 38 and product
1728.
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Answer / raghavendra sai
Let the three numbers in GP: a/r, a, ar---------(A)
Where '^' is power of . . .
Sum of these numbers are:
a/r +a +ar = 38
a(1\r+1+r) = 38 ---- (1)
Product of these numbers are:
a^3 = 1728
= (12)^3
a = 12
Putting the value of a in (1) you will get:
12(1\r+1+r) = 38
And factorising, we get
r = 2/3 or r = 3/2
Sub. the r and a value in (A), we get
8,12,18 or 18,12,8. When a = 12
And smallest no. is 8.
Read more: http://wiki.answers.com/Q/Find_the_smallest_number_in_a_GP_whose_sum_is_38_and_product_1728#ixzz1hFVTQBE3
| Is This Answer Correct ? | 27 Yes | 4 No |
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