Find the 10th and nth terms of the GP - 3/4,1/2, -1/3, 2/9 ......

Question

Find the 10^th and n^th terms of the GP \(-\frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9} .....\)

Answer 1

Given GP is \(-\frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9} .....\)

The given GP is of the form, a, ar, ar² , ar³…. 

Where r is the common ratio.

The first term in the given GP, a = a₁ = \(\frac{-3}{4}\)

The second term in GP, a₂ = \(\frac{1}{2}\)

Now, the common ratio,  \(r = \frac{a_2}{a_1}\)

r = \(-\cfrac{\frac{1}{2}}{\frac{3}{4}}\) = \(-\frac{2}{3}\)

Now, n^th term of GP is, an = ar^{n – 1} 

So, the 10^th term, a₁₀ = ar₉

Now, the required nth term, aⁿ = ar^n-1

Hence, the 10^th term, a₁₀ = \(\frac{128}{6561}\) and n^th term,

a_n = \(\big(\frac{9}{8}\big)\)\(\big(\frac{2}{3}\big)^n\)