Question

6th term and 13th term of a gp are 64 and8192.Find the 10th term of gp

Answer 1

a=2 r=2 and 10 term of GP will be=2*2*2*2*2*2*2*2*2=1024

Answer 2

Answer:

6th term of GP = 64

⇒ a(r)ⁿ⁻¹ = 64

a(r)⁶⁻¹ = 64

a(r)⁵ = 64

a = 64/r⁵ — (i)

13th term of GP = 8192

⇒ a(r)ⁿ⁻¹ = 8192

a(r)¹³⁻¹ = 8192

a(r)¹² = 8192

a = 8192/r¹² — (ii)

Now, we compare equations (i) and (ii):-

Since we know that both the values stand for ‘a’, we can say that they are equal to each other.

8192/r¹² = 64/r⁵

8192r⁵ = 64r¹²

8192r⁵/64 = r¹²

128r⁵ = r¹²

128 = r¹²/r⁵

128 = r⁷

2 = r  [7th root of 128 is 2]

Now, we find the value of ‘a’ by substituting the value of ‘d’ in equation (i):-

a = 64/r⁵

a = 64/2⁵

a = 64/32

a = 2

Now to find the 10th term of the GP:-

Formula = a(r)ⁿ⁻¹

⇒ 2(2)¹⁰⁻¹

2(2)⁹

2 × 512

1024 Ans

The 10th term of GP = 1024

Hope it helps

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