can anyone explain this solution in words? it would help me out so much. thanks?

Question

Problem

In an arithmetic sequence of positive numbers, the common difference is twice the first term, and the sum of the first six terms is equal to the square of the first term. Find the first term of the sequence.


Solution

36.

If a is the first term, then 2a is the common difference between successive terms, so the sum of the first six terms is

a + (a + 2a) + (a + 4a) + (a + 6a) + (a + 8a) + (a + 10a) = 36a

which also equals a2, so a = 36 is the first term.

Answer 1

The only step missing is:

a + (a + 2a) + (a + 4a) + (a + 6a) + (a + 8a) + (a + 10a) = 36a
a + (a + 2a) + (a + 4a) + (a + 6a) + (a + 8a) + (a + 10a) = a^2
36a = a^2
36 = a

In English:
Part 1. You have a sequence of numbers, where the common difference is twice the first term, meaning that you're adding 2 times the first term to each successive term.

Therefore, the sequence is expressed as:
a + (a + 2a) + (a + 4a) ... + (a + 2(n-1)a) where n is the number of terms.
You can re-express it as:
a + 3a + 5a + 7a) ... + ((2n-1)a) where n is the number of terms.

Part 2. The sum of the first 6 terms is equal to the square of the first term.

To express this, write out the first 6 terms, and then combine like terms:
a + (a + 2a) + (a + 4a) + (a + 6a) + (a + 8a) + (a + 10a)
a + 3a + 5a + 7a + 9a + 11a
36a

which equals the square of the first term (a^2)
36a = a^2
36 = a

Answer 2

I don't get what u're asking!!!!!!!!
everyting is explained in what u wrote!!!!!!!!
& why did u put the sum of the 6 terms 36 while the Q said is a squared????