in an arithmetic sequence whose first term is 4, the 1st, 3rd and 7th terms form consecutive terms of geometric sequence, find the sum of the first three terms of the arithmetic sequence

a + a+d + a+2d = 3 a + 3 d = 3(a+d)=12+3d

firat = 4
third = 4+2d
7th = 4+6d
r
and the geometric is
b, br, br^2, br^3 ...
4 = b
4+2d = b r = 4r
4+6d = b r^2 = 4 r^2

so
2 d = 4 r - 4 or d = 2r-2
6 d = 4 r^2 -4 or d = (2/3)r^2 -2/3

2 r - 2 = (1/3) (r^2 - 1)

6 r - 6 = r^2 - 1

r^2 - 6 r + 5 = 0
(r-5)(r-1) = 0
r = 5 or 1
1 is boring, constant terms in geo sequence
look at r = 5
d = 2 r - 2 = 8
then finally
12+3d = 12 + 24 = 36

picking it up from Damon's

first = 4
third = 4+2d
7th = 4+6d

but these also form a GP

(4+2d)/4 = (4+6d)/(4+2d)
16 + 16d + 4d^2 = 16 + 12d^2
8d^2 -16d = 0
8d(d - 2) = 0
d = 0 , or d = 2

for a=4, and d=2, the sum of the first three AP terms are
4 + 6 + 8
= 18

check:
term1 = 4
term3 = a+2d = 8
term7 = a+6d = 16
and 4,8, and 16 form a GP

41

Well, it seems we have to put on our thinking caps for this one! So, let's break this down step by step, shall we?

First, let's find the common difference (d) of the arithmetic sequence. Since we're given the first term as 4, let's assume the next term is 4 + d.

Now, we know that the 1st, 3rd, and 7th terms form consecutive terms of a geometric sequence. In other words, if we divide the 3rd term by the 1st term, and divide the 7th term by the 3rd term, we should get the same ratio.

Let's set up an equation to solve this. We have:

(4 + 2d) / 4 = (4 + 6d) / (4 + 2d)

Let's simplify this equation. Cross multiplying and simplifying, we get:

4(4 + 6d) = (4 + 2d)(4 + 2d)

Simplifying further, we get:

16 + 24d = 16 + 8d + 8d + 4d^2

Simplifying again, we have:

16 + 24d = 16 + 16d + 4d^2

Rearranging the terms, we get:

4d^2 - 8d = 0

Factoring out 4d, we have:

4d(d - 2) = 0

Therefore, either d = 0 or d - 2 = 0. Since d cannot be 0 (that would make every term the same), we have d = 2.

Now that we have the common difference, let's find the first three terms of the arithmetic sequence:

First term = 4
Second term = 4 + 2(2) = 8
Third term = 4 + 2(4) = 12

Finally, let's find the sum of the first three terms:

4 + 8 + 12 = 24

So, the sum of the first three terms of the arithmetic sequence is 24. Ta-da!

To find the sum of the first three terms of the arithmetic sequence, we first need to determine the common difference (d) of the sequence.

Let's denote the arithmetic sequence as: 4, a, b, c, d, e, f, ...

Since the first term is 4, the second term (a) can be found by adding the common difference (d) to the first term: a = 4 + d.

Similarly, the third term (b) will be: b = a + d = a + 2d.

Given that the first, third, and seventh terms (4, b, and f respectively) form consecutive terms of a geometric sequence, we can set up the following equation:

b^2 = 4 * f.

Substituting the values of b and f, we get:

(a + 2d)^2 = 4 * (a + 6d).

Expanding the equation:

a^2 + 4ad + 4d^2 = 4a + 24d.

Simplifying the equation:

a^2 - 4a + 4d^2 - 20d = 0.

Now, we can use this equation to find the value of d by solving for a in terms of d:

a = 2 ± √(1 - d^2 + 5d).

Since we are dealing with an arithmetic sequence, a and d should be real numbers. This means that the discriminant (1 - d^2 + 5d) must be greater than or equal to zero.

Solving for the discriminant: 1 - d^2 + 5d ≥ 0.

This inequality can be factored as: (d - 1)(d + 5) ≥ 0.

The solutions to this inequality are: d ≤ -5 or d ≥ 1.

Thus, the common difference (d) can be any value less than or equal to -5 or any value greater than or equal to 1.

Now that we have determined the range of possible values for d, we can evaluate the sum of the first three terms of the arithmetic sequence for each case.

Case 1: d ≤ -5
Let's say d = -5, then the sequence becomes: 4, -1, -6, -11, -16, -21, ...

The sum of the first three terms is: 4 + (-1) + (-6) = -3.

Case 2: d ≥ 1
Let's say d = 1, then the sequence becomes: 4, 5, 6, 7, 8, 9, ...

The sum of the first three terms is: 4 + 5 + 6 = 15.

In summary, depending on the possible values of the common difference (d), the sum of the first three terms of the arithmetic sequence can be either -3 or 15.

In AP,a=4, 3rd=4+2d, 7th=4+6d. We are told that: 4=a in GP, 4+2d=4r in GP and 4+6d=4r^2. So it becomes: 4+2d=4r(¡) and 4+6d=4r^2(¡¡). So in (¡) d=2r-2, and substitute in (¡¡) and becomes: 4+6(2r-2)=4r^2, and becomes 4r^2-12r+8=0. Which is (r-2)(r-1)=0. So r =1 or 2 . Substituting in AP 1st term is 4 and 3rd term is 8, then suntract 4 from 8 and diide by 2 to find commom difference,which is 2. So AP terms are:4,6.8,...... So AP sum 1st 3terms= 4+6+8 =18. Sum using formula=n/2{2a+(n-1)d}, substituting it becomes 3/2(8+4)=18.