Find the 4th term from the end of the GP 2/27, 2/9, 2/3, ........ 162.

Question

Find the 4^th term from the end of the GP \(\frac{2}{27}\), \(\frac{2}{9}\), \(\frac{2}{3}\), ..... 162.

Answer 1

The given GP is \(\frac{2}{27}\), \(\frac{2}{9}\), \(\frac{2}{3}\), ...... 162. → (1)

The first term in the GP, a₁ = a = \(\frac{2}{27}\)

The second term in the GP, a₂ = \(\frac{2}{9}\)

The common ratio, r = 3 

The last term in the given GP is a_n = 162. 

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 r’ = \(\frac{1}{r}\).

So, r' = \(\frac{1}{3}\)

So, 4^th term of the GP (2),

a₄ = ar³

= 162 x \(\frac{1}{3^3} = 6\)

Hence, the 4^th term from the end of the given GP is 6.