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Sagot :
Answer:
Not a G.P
Step-by-step explanation:
Given - A G.P is such that the third term is nine times the first term, while the second term is one twenty fourth of the fifth term
To find - find its fourth term.
Proof -
Let first term of G.P = a
Second term of G.P = b
Third term of G.P = c
Fourth term of G.P = d
Fifth term of G.P = e
Now,
Given that the third term is nine times the first term
⇒c = 9a
And
Given that the second term is one twenty fourth of the fifth term
⇒b = [tex]\frac{1}{24}[/tex] e
We know that in a g.P, there is common ratio in the terms
i.e [tex]\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = \frac{e}{d}[/tex] = r
and the nth term is represented as aₙ = arⁿ⁻¹
Now,
Second term would be b = ar
Third term would be c = ar²
fourth term would be d = ar³
Fifth term would be e = ar⁴
Now,
c = 9a
⇒ar² = 9a
⇒r² = 9
⇒r = 3 or -3
And
b = [tex]\frac{1}{24}[/tex] e
⇒24b = e
⇒24(ar) = ar⁴
⇒24 = r³
As we have r = 3 or -3
S0 (3)³ = 27 ≠ 24
or (-3)³ = 27 ≠ 24
It is not possible.
So, it is not a G.P
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