If the first, third and thirteenth terms of an arithmetic

If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence?


Slightly different algebraic approach.........

the first, third and thirteenth terms of an arithmetic progression are in geometric progression

First term = a
Third Term = a+2d
Thirteenth Term = a+12d

They are in Geometric Progression too
So \(\frac{(a+2d)}{a} = \frac{(a+12d)}{(a+2d)}\) --------------------------> {When a, b, c, d are in GP then b/a = c/b = d/c.........}

\((a+2d)^2 = a(a+12d)\) ----------> \(a^2 + 4d^2 + 4ad = a^2 + 12ad\) -----------> d^2 = 2ad -----------> d(d-2a)=0---------> Statement 1

the sum of the fourth and seventh terms of this arithmetic progression is 40 -------> (a+3d) + (a+6d) = 40 --------> 2a + 9d = 40 --------> \(d = \frac{40 - 2a}{9}\)

Putting \(d = \frac{40 - 2a}{9}\) in the Statement 1, we get \((\frac{40 - 2a}{9})\)\((\frac{40 - 2a}{9} - 2a)\) -------------------> \((\frac{40 - 2a}{9})\)\((\frac{40 - 20a}{9})\) ----------> \(\frac{(40-2a)(40-20a)}{81} = 0\) ------------------> Either (40-2a)=0 ------> a = 20 or (40-20a)=0 -----> a = 2, which is the answer