4
$\begingroup$

The first term is $8$ and the common difference is $d\neq0$. The first, fifth, and eighth terms of the progression are the first, second, and third terms, respectively, of a geometric progression whose common ratio is $r$.

  • What are two equations connecting $d$ and $r,$ and how can we use this to show that $r=3/4$ and find the value of $d?$

  • What is the sum to infinity of the geometric progression?

  • How can we find the sum of the first eight terms of the arithmetic progression?

$\endgroup$
3
  • 5
    $\begingroup$ Hard to read, at least $\endgroup$
    – leonbloy
    Jun 16, 2014 at 0:41
  • $\begingroup$ Once you've looked at the posted hint for your previous question on this general topic, perhaps you can tell us what your thoughts are on this problem. $\endgroup$
    – rogerl
    Jun 16, 2014 at 0:42
  • $\begingroup$ You should also show your own work, after you get the answer to the first you can solve the rest also $\endgroup$
    – Cloverr
    May 9, 2016 at 13:25

5 Answers 5

3
$\begingroup$

The series starts with a, and increases with d, so you get the numbers to be

$a,a+d,a+2d,a+3d,a+4d,a+5d,a+6d,a+7d$ as your 8 terms.

$a, a+4d, a+7d$ are your terms in your geometric one. The multiplication from the first two terms is $\frac{a+4d}{a} $ or $ 1 +\frac{4d}{a}$, and the second multiplications is $\frac{a+7d}{a+4d} $ or $ 1 +\frac{3d}{a+4d}$

For a geometric progression, the multiplication is always the same so $$ 1 +\frac{4d}{a} =1 +\frac{3d}{a+4d}$$ $$ \frac{4d}{a} =\frac{3d}{a+4d}$$ $$ 4d =\frac{3d*a}{a+4d}$$ $$ 4da+ 16d^2 =3da$$ $$ da+ 16d^2 =0$$ $$ 16d^2 =-da$$ $$ -16d =a$$

This means that we can rewrite those geometric progression ratios to

$$ 1 +\frac{4d}{-16d} $$ $$ 1 +-\frac{1}{4} = 3/4$$

$\endgroup$
2
$\begingroup$

The first, fifth, and eight terms of the arithmetic progression are $$ a, \quad a+4d, \quad a+7d. $$ The first three terms of the geometric progression are $$ a, \quad ar, \quad ar^2. $$ So we have \begin{align} a+4d & = ar, \\ a+7d & = ar^2. \end{align} Subtracting $a$ from both sides of both equations and then doing some routine algebra, we get \begin{align} 4d & = a(r-1), \\ 7d & = a(r^2-1) = a(r-1)(r+1). \end{align} Dividing the two sides of the second equation respectively by the two sides of the first, we get $$ \frac 7 4 = r+1, $$ so $r=\dfrac 3 4$. Hence $a+4d=3a/4$, and so $d=-a/16$.

$\endgroup$
0
$\begingroup$

We can write the arithmetic progression as $8, 8+d, 8+2d,\ldots$ and the geometric progression as $a, ar, ar^2,\ldots$

We know that the "the first term, the fifth term, and the eighth term of the [arithmetic] progression are the first term, the second term and the third term, respectively, of a geometric progression".

The equations that follow from these facts are sufficient to determine the three variables.

You can then use the standard formulae to determine the infinite sum of the geometric series and the partial sum of the arithmetic series of the first $8$ terms.

$\endgroup$
0
$\begingroup$
  1. List item

A.P in the given problem is 8,8+d,8+2d...

And it is given that the 1st,5th and 8th terms of the A.P are equal to the

,2nd and 3rd terms of a G.P.

First step.

First term of A.P is equal to the first term of G.P.

So, t1=a1=8

Second step.

t5(fifth term of A.P)=a2(second term of G.P)

i.e ar=t+4d

a=t=8 So,

8r=8+4d

2r=2+d after dividing both sides by 4

This is the equation which is the relation between ratio and the common difference.

Let this be eq1.

Third step.

a3=t8 as per the given information

i.e ar.r=t+7d where a=t=8

So, 8r.r=8+7d

as r=2+d/2 as per the eq1

we get 8r.2+d/2=8+7d

i.e 8r+4dr=8+7d

8r-8=7d-4dr

8.d+2/2-8=7d-4dr

4d+8-8-7d=-4dr

-3d=-4dr after cancelling -d on both sides we get

3=4r

and finally r=3/4, so the common ratio is 3/4 in the G.P By substituting this value in eq1 we get the value of d which is -1/2.

And by applying the formulae for sum to infinity of the G.P and sum of first 8terms of the A.P which are 16/3 and 9/2 respectively.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – AugSB
    May 9, 2016 at 13:28
0
$\begingroup$

WLOG let the 1st, 5th and 8th terms of the AP be $a-4d, a, a+3d$.

As these are also the 1st, 2nd and 3rd terms respectively of a GP, the common ratio of the GP is

$$r=\frac {\overbrace{\;\;\;a\;\;\;}^P}{\underbrace{a-4d}_R}= \frac {\overbrace{a+3d}^{Q}}{\underbrace{\;\;\;a\;\;\;}_S} =\color{red}{\frac {\overbrace{\;\;\;3\;\;\;}^{Q-P}}{\underbrace{\;\;\;4\;\;\;}_{S-R}}}$$ by using componendo and dividendo.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .