Final answer:
The lowest number in the AP is 15/7.
Step-by-step explanation:
To find the lowest number in an arithmetic progression (AP) such that the sum of all the terms is 105 and the greatest term is 6 times the least, we can use the formula:
Sum = (n/2)(first term + last term)
In this case, the sum is 105 and the ratio between the first term and the last term is 1:6. Let the first term be 'a' and the last term be '6a'.
Using the formula, we have:
105 = (n/2)(a + 6a)
105 = (n/2)(7a)
15 = 7an
an = 15/7
From here, we need to find the lowest value of 'n' that gives us an integer value for 'a'. We find that 'n = 7' and 'a = 15/7' satisfy this condition. Therefore, the lowest number in the AP is 15/7.