The positive numbers e and f are the 2nd and 14th terms respectively of an arithmetic progression whose first term is 1.It is also given that e,9 and f are three consecutive terms of a geometric progression. Find the values of e and f. ?

Question

Liam Brennan · Accepted Answer

The values are #e=3# and #f=27#

For the arithmetic progression, let the common ratio #=d#

Then,

#{(a_1=1),(a_2=e=1+d),(a_14=f=1+13d):}#

For the geometric progression, let the common ratio #=r#

Then,

#{(u_1=e),(u_2=er=9),(u_3=er^2=f):}#

From those #6# equations, we deduce that

#e=1+d=9/r#

#f=1+13d=9r#

Therefore,

#r=9/(1+d)=(1+13d)/9#

#=>#, #81=(1+13d)(1+d)#

#=>#, #13d^2+14d+1=81#

#=>#, #13d^2+14d-80=0#

Solving this quadratic equation in #d#

#d=(-14+-sqrt(14^2-4(13)(-80)))/(2*13)#

#d=(-14+-66))/(26)

Keeping only the positive value

#d=(66-14)/26=2#

Therefore,

#e=3#

and

#f=1+13*2=27#

Ariana Alvarez · Answer

undefined To find the values of $ e $ and $ f $, we can use the given information about arithmetic and geometric progressions.

Given that $ e $ and $ f $ are the 2nd and 14th terms, respectively, of an arithmetic progression with a first term of 1, we can use the formula for the nth term of an arithmetic progression:

$$ e = 1 + (2-1)d $$
$$ f = 1 + (14-1)d $$

Where $ d $ is the common difference of the arithmetic progression.

Next, we're given that $ e $, 9, and $ f $ are three consecutive terms of a geometric progression. Therefore:

$$ f = e \times r $$
$$ 9 = e \times r^2 $$

Where $ r $ is the common ratio of the geometric progression.

Now, we can solve the system of equations:

$$ e = 1 + (2-1)d $$
$$ f = 1 + (14-1)d $$
$$ f = e \times r $$
$$ 9 = e \times r^2 $$

Solving these equations will give us the values of $ e $ and $ f $.