SOLUTION: The sum of the first three terms of a geometric sequence of integers is equal to seven times the first term, and the sum of the first four terms is $30$. What is the first term of

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Question 1209793: The sum of the first three terms of a geometric sequence of integers is equal to seven times the first term, and the sum of the first four terms is $30$. What is the first term of the sequence?
Found 3 solutions by greenestamps, Edwin McCravy, math_tutor2020:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The sum of the first three terms of a geometric sequence of integers is equal to seven times the first term.

By inspection, the first three terms are a, 2a, and 4a. So the common ratio is 2.

The sum of the first four terms is 30.

The 4th term is 8a; the sum of the first four terms is a+2a+4a+8a = 15a. 15a = 30, so a = 2.

The sequence is 2, 4, 8, 16, ....

ANSWER: The first term is 2

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

Your teacher might expect you to use formulas and factoring of the sum or 
difference of cubes.  If so:

The sum of the first three terms of a geometric sequence of integers 

The formula for the first n terms of a geometric sequence is:
S%5Bn%5D=%28a%5B1%5D%28r%5En-1%29%29%2F%28r-1%5E%22%22%29

So the first 3 terms of a geometric sequence is

S%5B3%5D=%28a%5B1%5D%28r%5E3-1%29%29%2F%28r-1%5E%22%22%29

is equal to seven times the first term,  

So 
%28a%5B1%5D%28r%5E3-1%29%29%2F%28r-1%5E%22%22%29%22%22=%22%227%2Aa%5B1%5D
Dividing both sides by a1,

%28r%5E3-1%29%2F%28r-1%5E%22%22%29%22%22=%22%227

r%5E3-1%22%22=%22%227%28r-1%29

Factor the left side as the difference of two cubes:

%28r-1%29%28r%5E2%2Br%2B1%29%22%22=%22%227%28r-1%29

Dividing both sides by r-1:

r%5E2%2Br%2B1%22%22=%22%227

Adding -7 to both sides:

r%5E2%2Br-6%22%22=%22%220

Factoring the left side

%28r%2B3%29%28r-2%29%22%22=%22%220

r+3 =  0;   r-2 = 0
  r = -3;     r = 2

and the sum of the first four terms is $30$.  

S%5B4%5D=%28a%5B1%5D%28r%5E4-1%29%29%2F%28r-1%5E%22%22%29

So using r = -3  

%28a%5B1%5D%28%28-3%29%5E4-1%29%29%2F%28%28-3%29-1%5E%22%22%29%22%22=%22%2230

%28a%5B1%5D%2881-1%29%2F%28-4%29+%29%22%22=%22%2230

a%5B1%5D%2880%2F%28-4%29%29%22%22=%22%2230

a%5B1%5D%28-20%29%22%22=%22%2230

a%5B1%5D%22%22=%22%2230%2F%28-20%29

a%5B1%5D%22%22=%22%22-3%2F2

But this cannot be the answer, because this is a geometric 
sequence of integers and that is not an integer.

So using r = 2  

%28a%5B1%5D%28%282%29%5E4-1%29%29%2F%28%282%29-1%5E%22%22%29%22%22=%22%2230

a%5B1%5D%28%2816-1%29%2F1%29%22%22=%22%2230

a%5B1%5D%2815%29%22%22=%22%2230

a%5B1%5D%22%22=%22%2230%2F15

a%5B1%5D%22%22=%22%222

What is the first term of the sequence? 

Answer = 2

The sequence is 2, 4, 8, 16 and the first three terms sum to 2+4+8=14,
and 14 is 7 times the first term 2.

The sum of the first four terms is 2+4+8+16 = 30.  So this is correct.

Edwin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

a = first term
r = common ratio
ar = second term
ar*r = ar^2 = third term

sum of first 3 terms = 7*(first term)
a+ar+ar^2 = 7a
a(1+r+r^2) = 7a
1+r+r^2 = 7
r^2+r-6 = 0
(r+3)(r-2) = 0
r+3 = 0 or r-2 = 0
r = -3 or r = -2

If r = -3, then we have this geometric sequence
a, -3a, 9a, -27a
Those first four terms add to 30
a+(-3a)+9a+(-27a) = 30
-20a = 30
a = 30/(-20)
a = -1.5
But the instructions mention that each term of the geometric sequence is an integer.
We must rule out the r = -3 case.

If instead r = 2, then,
a+2a+4a+8a = 30
15a = 30
a = 30/15
a = 2 is the first term.