Geometric and Arithmetic Progressions?

Question

The $1st, 2nd and 3rd$ terms of a geometric progression are the $1st, 9th, and 21st$ terms respectively of an arithmetic progression. The $1st$ term of each progression is $8$ and the common ratio of the geometric progression is $r$ , where $r != 1$ . Find
(i) the value of r,
(ii) the $4th$ term of each progression.

1591 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-31T19:03:13

$(i) : r=3/2, (ii) :" the "4^(th)" term of GP is "27" and that of AP is "25/2$ .

The $1^(st)$ term of the GP is $8$ , and the common

ratio is $r!=1$ .

Hence, the $2^(nd)$ and the $3^(rd)$ terms of the GP must be

$8r$ and $8r^2$ .

Now, it is given that these are $9^(th) and 21^(st)$ terms of an AP, of

which, $1^(st)$ term is $8$ .

So, if $d$ is the common difference of the AP, then, we have

$8r=8+(9-1)d, i.e., 8r=8+8d, or, r=1+d...(ast^1)$ ,

$8r^2=8+20d, or, 2r^2=2+5d...................(ast^2)$ .

$"By "(ast^1)" then, "2r^2=2+5(r-1)=5r-3$ .

$rArr 2r^2-5r+3=0$ .

$rArr (r-1)(2r-3)=0$ .

$:. r=1$ , which contradicts the given that $r!=1, or, r=3/2$ .

Therefore, the $4^(th)$ of the GP is $ar^3=8*(3/2)^3=27$ , and, that of the AP is $8+(4-1)d=8+(4-1)(3/2)=25/2$ .