Question
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.
Answers
oobleck
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.
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.
Bosnian
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
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
oobleck
nicely done, @Bosnian. I guess I'll have to take a second look at my analysis.