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Find the first 4 terms of a G.P. if a = -4 and r = -2.

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- 4, 8, - 16, 32, \dots

4, 8, 16, 32, \dots

- 4, - 8, - 16, - 32, \dots

4, 8, - 16, 32, \dots

answer is A.

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Detailed Solution

A finite geometric progression or GP sequence in terms of first term a and common ratio r for n terms can be written as a,

{ar}^{2}, {ar}^{3}

{ar}^{4}

,...,

{ar}^{n - 1}

. We see that the given question does not mention the number of terms, so it will be infinite GP as a,

{ar}^{2}, {ar}^{3}

{ar}^{4}

,…..
Mathematically, a sequence with infinite terms is written as
(

\times_{n}

) =

\times_{1} {, \times}_{2},

\times_{3}

,…..
If, a sequence with finite terms is written as
(

\times_{n}

) =

\times_{1} {, \times}_{2},

\times_{3}

,…..

\times_{n}

GP is a type sequence where the ratio between any two consecutive numbers is constant. If (

\times_{n}

) =

\times_{1} {, \times}_{2},

\times_{3}

,….. is an GP, then

\frac{\times_{2}}{\times_{1}}

\frac{\times_{3}}{\times_{1}}

= r………. (i)
Here the ratio between two terms is called the common ratio and denoted as r. The first term is conventionally denoted as a. We put

\times_{1}

= a in equation (1),
r =

\frac{\times_{2}}{\times_{1}} = \frac{\times_{2}}{a}

⇒

\times_{2}

= ar
We can similarly put

\times_{2}

= ar in equation (i),
r =

\frac{\times_{3}}{\times_{2}} = \frac{\times_{2}}{ar}

⇒

\times_{3}

= ar × r = ar2
Similarly, x4 = ar3, x5 = ar4…. We observe that the nth term of GP sequence is arn-1 and we can write the infinite GP sequence in terms of first term a and common ratio r as
a, ar2, ar3, ar4,…..
Given that a = −4 and r = −2
a = −4,
ar2 = (−4)( −2)2 = 8,
ar3 = (−4)( −2)3 = −16,
ar4 = (−4)( −2)4 = 32.
The first 4 terms of a G.P. if a = -4 and r = -2 are