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The sum of three numbers in G.P. is 70 . If the two extemes by multiplied each by 4 and the mean by 5,the products are in A.P. . The numbers are?
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The sum of three numbers in G.P. is 70 . If the two extemes by multipl...
Given Information:
- Let the three numbers in geometric progression be a/r, a, ar with common ratio r.
- Sum of the three numbers: a/r + a + ar = 70
- New numbers after multiplication: 4a/r, 5a, 4ar
- The products of the new numbers are in arithmetic progression.

Finding the Numbers:

Sum of the numbers:
a/r + a + ar = 70
Using the formula for the sum of a G.P., we get:
a(1/r + 1 + r) = 70
a(r^2 + 1 + r^2) = 70
2ar^2 + a = 70
a(2r^2 + 1) = 70
a = 70 / (2r^2 + 1)

Products of the new numbers:
4a/r * 4ar = 16a^2 = a(5a + a)
5a * 4ar = 20a^2 = a(4a + 5a)
The products are in A.P. if the two middle terms are twice the first term:
20a^2 = 2 * 16a^2
20a^2 = 32a^2
Solving the above equation, we get:
a = 0 (reject), a = 35

Therefore, the numbers are:
a/r = 35/r
a = 35
ar = 35r

Explanation:
- We first express the numbers in G.P. and find the sum to get the relation between the numbers.
- Then we determine the new numbers after multiplication and set up the condition for the products to be in A.P.
- Solving the equations, we find the value of 'a' which gives us the numbers in the G.P.
- Finally, we express the numbers in terms of 'a' and 'r' to get the final numbers.
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The sum of three numbers in G.P. is 70 . If the two extemes by multiplied each by 4 and the mean by 5,the products are in A.P. . The numbers are?
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