Question

if the third and the seventh terms of a GP are 4 and 64 respectively find the tenth term.


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Answer 1

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g₃ = 4

g₇ = 64

g₇/g₃ = 64/4

g₇ = ar⁷⁻¹ = ar⁶

g₃ = ar³⁻¹ = ar²

so, $\frac{ar^6}{ar^2} = \frac{64}{4}$

=> r⁶⁻² = 16

=> r⁴ = 16

=> r = 2

ar² = 4

a x 2² = 4

a = 4/4 = 1

g₁₀ = ar¹⁰⁻¹ = ar⁹ = 1 x 2⁹ = 512

Answer 2

Step-by-step explanation:

No worries, I can help you with that!

Let the first term of the GP be "a" and the common ratio be "r".

Then, the third term is ar^2 = 4, and the seventh term is ar^6 = 64.

Dividing the two equations, we get:

(ar^6) / (ar^2) = 64/4

Simplifying, we get:

r^4 = 4

Taking the fourth root of both sides, we get:

r = ±√2

Since we want a positive common ratio, we take r = √2.

Now we can use the formula for the nth term of a GP:

an = ar^(n-1)

Substituting n = 10, a = ?, and r = √2, we get:

a10 = a√2^9

We still need to find the value of a. To do this, we can use the third term of the GP:

a√2^2 = 4

a = 4 / 2 = 2

Now we can substitute a = 2 into the formula for the tenth term:

a10 = 2√2^9

a10 = 512

Therefore, the tenth term of the GP is 512.