Question

If the fourth term of a G.P. is 2, then what is the product of its first 7 terms?​

Answer 1

Answer:

answer for the given problem is given

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Seven consecutive terms of an Geometric Progression can be taken as

$\displaystyle \: \frac{a}{ {r}^{3} } \: , \: \frac{a}{ {r}^{2} } \: , \: \frac{a}{ {r}^{} } \: , \: a \: , \: {a}{r} \: , \: {a}{ {r}^{2} } \: , \: {a}{ {r}^{3} }$

GIVEN

Fourth term of a Geometric Progression is 2

TO DETERMINE

The product of its first 7 terms of the Geometric Progression

EVALUATION

Let Seven consecutive terms of an Geometric Progression are

$\displaystyle \: \frac{a}{ {r}^{3} } \: , \: \frac{a}{ {r}^{2} } \: , \: \frac{a}{ {r}^{} } \: , \: a \: , \: {a}{r} \: , \: {a}{ {r}^{2} } \: , \: {a}{ {r}^{3} }$

So by the given condition

$a = 2$

RESULT

Hence the required product of its first 7 terms of the Geometric Progression

$= \displaystyle \: \frac{a}{ {r}^{3} } \: \times \: \frac{a}{ {r}^{2} } \: \times \: \frac{a}{ {r}^{} } \: \times \: a \: \times \: {a}{r} \: \times \: {a}{ {r}^{2} } \: \times \: {a}{ {r}^{3} }$

$= {a}^{7}$

$= {2}^{7}$

$= 128$