Question

The third term of a G.P. is greater than its first term by 9 whereas its second term is greater than the fourth term by 18. Find the G.P.

Answer 1

Let first term be a and common ratio be r.

By formula,

⇒ a_n = ar^{n − 1}

Given,

The third term of a G.P. is greater than its first term by 9.

∴ a₃ − a = 9

⇒ ar^{3 − 1} − a = 9

⇒ ar² − a = 9

⇒ a(r² − 1) = 9

⇒ a = `9/(r^2−1)`   .......(1)

Given,

Second term is greater than the fourth term by 18.

∴ a₂ − a₄ = 18

⇒ ar^{2 − 1} − ar^{4 − 1} = 18

⇒ ar − ar³ = 18

⇒ ar(1 − r²) = 18

Substituting value of a from equation (1) in above equation, we get:

⇒ `9/(r^2 − 1)​ × r(1 − r^2) = 18`

⇒ `9/(r^2 − 1) ​×  −r(r^2 − 1) = 18`

⇒ −9r = 18

⇒ r = `−18/9` ​

∴ r = −2

Substituting value of r in equation (1), we get:

⇒ `a = 9/((−2)^2−1)` 

= `9/(4−1)`

= `9/3`

= 3

G.P. = a, ar, ar², ar³, .......

= 3, 3 × −2, 3 × (−2)², 3 × (−2)³, .......

= 3, −6, 3 × 4, 3 × −8, ........

= 3, −6, 12, −24, ........

Hence, required G.P. = 3, −6, 12, −24, ........