Question

in a gp .the first term is 5 and the last term is 320 and the sum is 635. find the 4 the term​

Answer 1

Answer:

t4 = 40

Step-by-step explanation:

a = 5

tn = 320

Sn = 635

t4 = ?

let common ratio = r

now

tn = a(r)^n-1

=> 320 = 5 (r)^n-1

=> 320r = 5(r)^n ....(i)

now

Sn = a[(r)^n - 1] / r - 1

=> 635 = 5[(r)^n - 1] / r - 1

=> 635 = [5(r)^n - 5] / r - 1

=> 635 = [320r - 5] / r - 1 { from eq (i)

=> 635(r-1) = 320r - 5

=> 635r - 635 = 320r - 5

=> 315r = 630

=> r = 2

now

t4 = a[(r)^4 - 1]

=> t4 = 5[(2)^4 - 1]

=> t4 = 5(2)^3

=> t4 = 5 × 8

=> t4 = 40