The first,third and seventh term of an arithmetic progression form three consecutive terms of a geometric progression. If the first, second terms of the arithmetic progression is 6. Find its first term and common difference.
If your sentence means:
If the first of the arithmetic progression is 6.
then
In A.P.
an = a1 + ( n - 1 ) d
where
a1 = the initial term
d = common difference
an = the nth term
a3 = a1 + ( 3 - 1 ) ∙ d = a1 + 2 d
a7 = a1 + ( 7 - 1 ) ∙ d = a1 + 6 d
The members of G.P. are:
a1 , a3 , a7
Common ratio of G.P. is the quotient of two adjacent members of G.P.
r = a3 / a1 = ( a1 + 2 d ) / a1
r = a7 / a3 = ( a1 + 6 d ) / ( a1 + 2 d )
r = r
( a1 + 2 d ) / a = ( a1 + 6 d ) / ( a1 + 2 d )
Cross multiply.
( a1 + 2 d )² = a1 ∙ ( a1 + 6 d )
a1² + 2 ∙ a1 ∙ 2 d + ( 2 d )² = a1² + 6 a1 ∙ d
a1² + 4 a1 ∙ d + 4 d² = a1² + 6 a1 ∙ d
Subtract a1² to both sides.
4 a1 ∙ d + 4 d² = 6 a1 ∙ d
Subtract 4 a1 ∙ d to both sides.
4 d² = 2 a1 ∙ d
Divide both sides by 4
d² = 2 a1 ∙ d / 4
d² = a1 ∙ d / 2
Divide both sides by d
d = a1 / 2
Since a1 = 6
d = 6 / 2 = 3
a3 = a1 + 2 d
a3 = 6 + 2 ∙ 3 = 6 + 6 = 12
a7 = a1 + 6 d
a7 = 6 + 6 ∙ 3 = 6 + 18 = 24
Common ratio of this G.P is:
r = a3 / a1 = 12 / 6 = 2
Or, which is the same thing:
r = a7 / a3 = 24 / 12 = 2
A.P:
6 , 6 + 3 , 6 + 2 ∙ 3 , 6 + 3 ∙ 3 , 6 + 4 ∙ 3 , 6 + 5 ∙ 3 , 6 + 6 ∙ 3
6 , 9 , 12 , 15 , 18 , 21 , 24
first term = 6
third term = 12
seventh term = 24
G.P:
6 , 12 , 24
just use your usual formulas, and if the two sequences are
a, ar, ar^2, ...
b, b+d, B=2d, ...
then you have
a = b
ar^2 = b+d
ar^6 = b+2d
so now you have
ar^2 = a+d
ar^6 = a+2d
Now, I don't know what "the first, second terms of the arithmetic progression is 6" is supposed to mean, but I guess it means either of
the first term of the arithmetic progression is 6
the sum of first, second terms of the arithmetic progression is 6
So either
ar^2 = a+6
ar^6 = a+12
r^4 + r^2 + 1 = 2
no integer solutions there..
or,
ar^2 = a+d
ar^6 = a+2d
2a+d = 6
again, no easy solutions.
So I suspect some kind of typo.
nicely done, @Bosnian. I guess I'll have to take a second look at my analysis.
To solve this problem, we can use the properties of arithmetic and geometric progressions.
Let's denote the first term of the arithmetic progression as a and the common difference as d.
The first term of the arithmetic progression is given as 6. Hence, a = 6.
Now, let's find the third term of the arithmetic progression. Using the formula for the nth term of an arithmetic progression, we have:
a + (n - 1) * d
In this case, since the second term is a + d, the third term will be a + 2d.
So, a + 2d = 6 + 2d.
Now, we are told that the first, third, and seventh terms of the arithmetic progression form three consecutive terms of a geometric progression. In a geometric progression, the ratio between consecutive terms is constant.
Let's consider the ratio between the third and first terms of the arithmetic progression. We have:
(a + 2d)/a
Similarly, let's consider the ratio between the seventh and third terms of the arithmetic progression:
(a + 6d)/(a + 2d)
Since these ratios are constant, we can equate them:
(a + 2d)/a = (a + 6d)/(a + 2d)
Cross-multiplying and simplifying, we get:
(a + 2d)^2 = a(a + 6d)
Expanding and simplifying further, we have:
a^2 + 4ad + 4d^2 = a^2 + 6ad
Canceling out the common terms and rearranging, we get:
2ad + 4d^2 = 0
Factoring out d, we have:
2d(a + 2d) = 0
This equation gives us two possibilities: d = 0 or a + 2d = 0.
Since the common difference of an arithmetic progression cannot be zero (otherwise, it would not be an arithmetic progression), we have a + 2d = 0.
Substituting the value of a from earlier (a = 6), we can solve for d:
6 + 2d = 0
2d = -6
d = -3
Therefore, the first term (a) of the arithmetic progression is 6, and the common difference (d) is -3.