Q14 of 176 Page 21

In a GP the 3^rd term is 24, and the 6^th term is 192. Find the 10^th term.

T_n = ar^n-1

a = a, r =?, T_n = 24 n = 3

a = a, r =?, T_n = 192 n = 6

∴ 24 = a.r^3-1

⇒ 24 = a.r²…(1)

∴ 192 = a.r^6-1

⇒ 192 = a.r⁵…(2)

Divide (2) by (1) we get

⇒

⇒ r³ = 8

⇒ r = 2

Substituting r in 2 we get

a = 6

T₁₀ = 6.2^10-1

= 6.2⁹

= 3072.

If 5^th, 8^th and 11^th terms of a G.P. are p, q and s respectively, prove that a² = ps.

The 4th term of a G.P. is square of its second term, and the first term is -3. Find its 7th term.

If a, b, c, d and p are different real numbers such that :

(a² + b² + c²)p² – 2(ab + bc + cd) p + (b² + c² + d²) ≤ 0, then show that a, b, c and d are in G.P.

If then show that a, b, c, and d are in G.P.

In a GP the 3rd term is 24, and the 6th term is 192. Find the 10th term.