Question

If the product of the first four consecutive terms of a g.p is 256 and if the common ratio is 4 and the first terms is positive, then its 3rd terms is what

Answer 1

$Let \: a \: and \: r \: are \: first \:term \:and \\common \:ratio \: of \: a \: G.P$

$r = 4$

$Product \: of \: the \: first \:four \: terms = 256$

$\implies a \times ar \times ar^{2}\times ar^{3} = 256$

$\implies a^{4} \times r^{6} = 4^{4}$

$\implies a^{4} \times 4^{6} = 4^{4}$

$\implies a^{4} = \frac{4^{4} }{4^{6}}$

$\implies a^{4} = \frac{1}{4^{2}}$

$\implies a^{4} = \frac{1}{(2^{2})^{2}}$

$\implies a^{4} =\Big( \frac{1}{2}\Big)^{4}$

$\implies a= \frac{1}{2}\: [ a > 0 ]$

$Now, 3^{rd} \: term \: in \: G.P = ar^{2} \\= \frac{1}{2} \times 4^{2} \\= \frac{16}{2} \\= 8$

Therefore.,

$\red{ First \:term } \green { = \frac{1}{2}}$

$\red { 3^{rd} \: term \: in \: G.P}\green {= 8}$

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Answer 2

Answer:

8

Step-by-step explanation:

four consecutive numbers

a, ar, ar^2 , ar^3

their product a×ar×ar^2×ar^3=256

a^4 × r^6 = 256

a^4= 256/r^6

=256/4^6 (r=4)

=2^8/2^12

=(1/2)^4

a =1/2

t3 =ar^2

= 1/2 × 4^2

=1/2 × 8

t3=8