Find the 6th term from the end of GP 8, 4, 2… 1/1024

Question

Find the 6^th term from the end of GP 8, 4, 2… \(\frac{1}{1024}\)

Answer 1

The given GP is 8, 4, 2… \(\frac{1}{1024}\).→ (1)

First term in the GP, a₁ = a = 8 

Second term in the GP, a₂ = ar = 4

The common ratio, r = \(\frac {4}{8}\) = \(\frac{1}{2}\)

The last term in the given GP is \(\frac{1}{1024}\).

Second last term in the GP = a^n-1 = ar^n-2

Starting from the end, the series forms another GP in the form,

ar^n-1 , ar^n-2 , ar^n-3….ar³ , ar² , ar, a → (2)

Common ratio of this GP is \(\frac{1}{r}\)

So, common ratio = 2

a = \(\frac{1}{1024}\)

So, 6^th term of the GP (2),

a₆ = ar⁵

= \(\frac{1}{1024}\) x 2⁵ = \(\frac{1}{32}\)

Hence, the 6th term from the end of the given GP is \(\frac{1}{32}\) .