Geometric and Arithmetic Progressions?

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Geometric and Arithmetic Progressions?

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

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Answer 1 · 2018-05-31T19:03:13

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

Explanation:

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

ratio is $r \neq 1$ $r!=1$ .

Hence, the $2^{n d}$ $2^(nd)$ and the $3^{r d}$ $3^(rd)$ terms of the GP must be

$8 r$ $8r$ and $8 r^{2}$ $8r^2$ .

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

which, $1^{s t}$ $1^(st)$ term is $8$ $8$ .

So, if $d$ $d$ 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 $8+(4-1)d=8+(4-1)(3/2)=25/2$ .

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Geometric and Arithmetic Progressions?

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