Question

find t5 and s4 of the geometric progression if the first term is 27/16 and the common ratio is 2/3

Answer 1

Given:

Find t5 and s4 of the geometric progression if the first term is 27/16 and the common ratio is 2/3

To find:

t5 and s4

Solution:

The first term of the G.P., a = $\frac{27}{16}$

The common ratio of the G.P., r = $\frac{2}{3}$

Finding the value of t₅:

The formula for the nth term of a G.P. is,

$\boxed{\bold{T_n = a( r)^n^-^1}}$

where Tₙ = last term, a = first term, r = common ratio and n = term position

∴ $t_5 = \frac{27}{16} \times (\frac{2}{3} )^5^-^1$

$\implies t_5 = \frac{27}{16} \times (\frac{2}{3} )^4$

$\implies t_5 = \frac{27}{16} \times \frac{16}{81}$

$\implies t_5 = \frac{27}{81}$

$\implies\boxed{ \bold{t_5 = \frac{1}{3}}}$

Finding the value of s₄:

The formula for the sum of n terms of a G.P. is,

$\boxed{\bold{S_n = \frac{a(r^n- 1)}{r - 1} }}$

where Sₙ = last term, a = first term, r = common ratio and n = term position

∴ $S_4 = \frac{\frac{27}{16}[(\frac{2}{3}) ^4- 1]}{\frac{2}{3} - 1}$

$\implies S_4 = \frac{\frac{27}{16}[\frac{16}{81}- 1]}{\frac{2}{3} - 1}$

$\implies S_4 = \frac{\frac{27}{16}[\frac{16- 81}{81}]}{\frac{2-3}{3}}$

$\implies S_4 = \frac{\frac{27}{16}\times \frac{-65}{81}}{\frac{-1}{3}}$

$\implies S_4 = \frac{27}{16}\times \frac{65}{81}}\times 3$

$\implies \boxed{\bold{S_4 = \frac{65}{16}}}$

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Also View:

the first term of a geometric progression is 81 and the common ratio is 2/6. the product of the first 10 terms is​?

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