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. ?

Answer 1

Liam Brennan

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#

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Answer 2

Ariana Alvarez

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 $f$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.