Question

in a g.p the second term is 2/3 and 5th term is 16/81. its 7th term is

Answer 1

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\mathcal{\bf{{\underline{\underline{\huge\mathcal{...Answer...}}}}}}$

$\underline{\large\mathcal\red{solution}}$

let the first term and the common ratio is

a and r

now .......

a(2)=ar^(2-1)=ar=2/3

a(5)=ar^(5-1)=ar^4=16/81

ar^4=16/81

ar×r^3=16/81

r^3=(16×3)/(81×2)=8/27

r=2/3

now ...a=1

a(7)=(1)(2/3)^(7-1)=(2/3)^6=64/729

the 7th term is 64/729

$\large\mathcal\red{hope\: this \: helps \:you......}$

Answer 2

Answer:

The seventh term is 64/729

Step-by-step explanation:

The general g.p series goes like this

a, ar, ar^2, ar^3......

To find a term in g.p series we need a and r

So,

1st term = a

2nd term = ar

3rd term = ar^2

Similarly, the 5th term = ar^4

the 7th term = ar^6

which implies,

ar = 2/3 ---- (i) and ar^4 = 16/81 ----(ii)

Now,

(ii) /(i) = (16/81) / (2/3)

ar^4 16 * 3

--------- = --------

ar 81 * 2

r^3 = 8/27

r^3 = (2/3)^3

r = 2/3 -----(iii)

Now, put (iii) in (i) to get the value of a

ar = 2/3

a = (2/3) / r

a = (2/3) / (2/3)

a = 1

Now,

seventh term = ar^6

= 1*(2/3)^6

= (2/3)^6

7th term = 64/729

I hope dis would help u!!