Verified Answer :

dalendrk

\(a_2=-12;\ a_5=768\\\\a_n=a_1r^{n-1}\\\\a_5:a_2=r^3\\\\r^3=768:(-12)\\\\r^3=-64\\\\r=\sqrt[3]{-64}\\\\r=-4\\\\a_1=a_2:r\to a_1=-12:(-4)=3\\\\\boxed{a_n=3\cdot(-4)^{n-1}}\)

Blacklash

Answer:

Nth term is given by \(a_n=3\times (-4)^{n-1}\)

Step-by-step explanation:

For a geometric progression we have expression for nth term

            \(a_n=ar^{n-1}\)

where a is first term and r is common ratio.

Here the second and fifth terms are -12 and 768.

That is

           a₂ = ar²⁻¹ = ar = -12

           a₅ = ar⁵⁻¹ = ar⁴ = 768

Dividing we will get

           \(\frac{ar^4}{ar}=\frac{768}{-12}\\\\r^3=-64\\r=-4\)

Substituting in

          ar = -12

          a x -4 = -12

          a = 3

Nth term is given by \(a_n=3\times (-4)^{n-1}\)

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