Question

The 3rd term of a GP is 2/3 and 6th term is 2/81 then the 1st term is​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{In a G.P}$

$\mathsf{t_3=\dfrac{2}{3}\;\&\;t_6=\dfrac{2}{81}}$

$\underline{\textsf{To find:}}$

$\textsf{First term of the G.P}$

$\underline{\textsf{Solution:}}$

$\underrline{\mathsf{Concept\;used:}}$

$\mathsf{The\;n\,th\;term\;of\;the\;G.P\;a,ar,ar^2.............\;is}$

$\boxed{\mathsf{t_n=ar^{n-1}}}$

$\mathsf{Consider,}$

$\mathsf{t_3=\dfrac{2}{3}}$

$\implies\mathsf{ar^2=\dfrac{2}{3}}$.....(1)

$\mathsf{t_6=\dfrac{2}{81}}$

$\implies\mathsf{ar^5=\dfrac{2}{81}}$.....(2)

$\mathsf{Divide\;(2)\;by\;(1)}$

$\mathsf{\dfrac{ar^5}{ar^2}=\dfrac{\dfrac{2}{81}}{\dfrac{2}{3}}}$

$\mathsf{r^3=\dfrac{2}{81}{\times}\dfrac{3}{2}}$

$\mathsf{r^3=\dfrac{1}{27}}$

$\implies\boxed{\mathsf{r=\dfrac{1}{3}}}$

$\text{From (1)}$

$\mathsf{a{\times}\dfrac{1}{9}=\dfrac{2}{3}}$

$\mathsf{a=\dfrac{2{\times}9}{3}}$

$\mathsf{a=2{\times}3}$

$\implies\boxed{\mathsf{a=6}}$

$\underline{\textsf{Answer:}}$

$\textsf{First term is 6}$

$\underline{\textsf{Find more:}}$

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If r=1 upon 3,a=9 find t7​

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Answer 2

Answer:

1st term is = 6

Step-by-step explanation:

general form of G.P is a, ar, ar², ar³, ar⁴,ar⁵... and so on

we have, 3rd term is = 2/3

6th term is = 2/81

we'll write it like G.P form :

ar² = 2/3

ar⁵ = 2/81

first we will find value of [r = ?]

and, then find value of [a = ?]

ar⁵/ar² = 2/81 × 3/2

after cross multiplication..we have

r³ = 1/27 , r = 1³/3³ , (r = 1/3)

now, we find value of [a = ?]

3rd term is 2/3

ar²= 2/3

a × (1/3)² = 2/3

a × 1/9 = 2/3

after cross multiplication we have..

3a = 18

(a = 6)..✌️