Name: Sum of Infinite Terms of G.P.
Uploaded: 2019-07-19
Duration: 3 min 36 s
Description: Do you know that we can find the sum of an infinite number of terms of G.P.? Of course, there will be some conditions. Check this video to learn more.

Question

The first, second and third terms of a geometric progression are the first, fourth and tenth terms, respectively, of an arithmetic progression. Given that the first term in each progression is $12$ and the common ratio of the geometric progression is $r$ , where $r \neq 1$ , find the sixth term of each progression.

Accepted Answer

Given that, the first, second and third terms of a geometric progression are the first, fourth and tenth terms, respectively, of an arithmetic progression.

Therefore, first term $a = 12$

Now, second term of geometric progression is tenth term of an arithmetic progression i.e., $a r = a + 3 d \to (1)$

Similarly, the third term of geometric progression is fourth term of an arithmetic progression i.e., $a r^{2} = a + 9 d \to (2)$

Now, apply $(2) - 3 \times (1)$ , then $a r^{2} - 3 a r = - 2 a$

$\Rightarrow r^{2} - 3 r + 2 = 0$

$\Rightarrow r^{2} - 2 r - r + 2 = 0$

$\Rightarrow r (r - 2) - 1 (r - 2) = 0$

$\Rightarrow (r - 1) (r - 2) = 0$

$\Rightarrow r = 1; 2$

Since, given that, $r \neq 1$

Hence, the required value of $r = 2$ .

Now, substitute the values of $a = 12; r = 2$ in $(1)$ , then $12 (2) = 12 + 3 d$

$\Rightarrow 3 d = 12$

$\Rightarrow d = 4$

Now, the sixth term of geometric progression is $t_{6} = a . r^{6 - 1} = 12 . (2^{5}) = 384$

Similarly, the sixth term of an arithmetic progression is $t_{6} = a + (6 - 1) d = 12 + 5 (4) = 12 + 20 = 32$ .

Hence, the sixth term of geometric progression is $384$ and the sixth term of an arithmetic progression is $32$ .

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