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?

A. 2
B. 3
C. 4
D. 5
E. 6

We can write \(a_1*a_{13}=a_3*a_3\) and \(a_3+a_7=40.\)
Since \(a_3=a_1+2d,\) (where \(d\) is the the constant defining the arithmetic progression), \(a_{13}=a_1+12d\), \(\, a_4=a_1+3d\), \(\, a_7=a_1+6d\), from the above two equations we get
\(a_1(a_1+12d)=(a_1+2d)^2\) and \(a_1+3d+a_1+6d=40.\) Further, \(12a_1d=4a_1d+4d^2\) or \(8a_1d=4d^2\) which leads to \(2a_1=d.\) (If \(d=0\) all the terms in the arithmetic progression are equal to 20, and so are those in the geometric progression. It should have been stated that the arithmetic progression is non-constant.)
The other equation leads to \(2a_1+9d=40\), so \(2a_1+18a_1=40,\) or \(a_1=2.\)

Answer A.

IMO, not a real GMAT question.
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That's wrong. The solution is: 2, 6, 10, 14, 18, 22, 26, ..., 50.
Geometric means constant factor, so in this case: 2, 10, 50 (constant factor being 5). And 14+26=40. It would be pretty hard to guess the correct solution. A lot of trial and error. It would take too long.
In your sequence, 16 isn't even the 13th term. It's the 15th.

exactly
16 will be 15th term not 13th
that itself can refute further calculation based on that approach
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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
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equation1: (a+2d)/a=(a+12d)/(a+2d)➡d=2a
equation2: 2a+9d=40
substituting,
2a+18a=40
a=2

Given 1st,3rd and 13th term in GP.
so \(b^{2}\)=ac (GP Mean rule)
\(t_3\)\(^{2}\)=\(t_1\)*\(t_{13}\)
\((a+2d)^{2}\)=a(a+12d)
\(a^{2}\)+4ad+4\(d^{2}\)=\(a^{2}\)+12ad
4\(d^{2}\)=8ad
4d(d-2a)=0 (d is not equal to zero)
d=2a------------- (1)

Given \(t_4\)+\(t_7\)=40
a+3d+a+6d=40
2a+9(2a)=40 from (1)
20a=40
First term a =2