In an arthmetic sequence of positive numbers the sommon difference is twice the first term and he sum of the first six terms is equal to the square of the first term. Find the first term in the sequence.
a_n = a_1 + (n – 1)d
where a_n is the current term, a_1 is the first term, d is the common difference.
Also note, that the sum of an arithmetic sequence is given by:
S_n = n(a_1 + a_n)/2 = n[2a_1 + (n – 1)d]/2
Now we have the common difference d = 2a_1, and the sum of the first 6 terms, S_6 = 6(a_1 + a_6)/2 = a_1^2
=> a_n = a_1 + (n – 1)2a_1 = a_1 + 2na_1 – 2a_1 = (2n – 1)a_1
=> a_6 = 11a_1
Now S_6 = a_1^2
=> 6(a_1 + a_6)/2 = a_1^2
=> 2a_1^2 = 6a_1 + 6a_6
=> 2a_1^2 = 6a_1 + 6(11a_1)
=> 2a_1^2 = 6a_1 + 66a_1
=> 2a_1^2 = 72a_1 …………………this is a quadratic equation in a_1, you can replace a_1 with a variable, say x, if all the symbols are getting you confused.
=> 2a_1^2 – 72a_1 = 0
=>2a_1(a_1 – 36) = 0
Since a_1 cannot be zero,
a_1 – 36 = 0
=> a_1 = 36
So the first term of the sequence is 36, and if you're interested, the sequence is a_n = 36 + (n – 1)72
That is 36, 108, 180, 252, 324, 396,…