Question

The third terms GP is 9 and the fifth term is 16.find the 4th term

Answer 1

Answer:

FOURTH TERM IS 12.

Step-by-step explanation:

$a(3) = 9 \: \: \: \: \: ...(i) \\ \\ a {r}^{2} = 9 \\ \\ a(5) = 16 \: \: \: \: \: ...(ii) \\ \\ a {r}^{4} = 16 \\ \\ solving \: these \: \\ \\ we \: get \\ \\ {a}^{2} {r}^{6} = 16 \times 9 \\ \\ {(a {r}^{3} )}^{2} = 144 \\ \\ a {r}^{3} = 12$

Answer 2

Given :-

• Third term of GP = 9
• Fifth term of GP = 16

To Find :-

• Fourth term of GP = ?

Formula used :-

• nth term of GP = a•r^(n - 1)
• a = First term
• r = common ratio .

Solution :-

Let us assume that, first term of given GP series is a and common ratio is r .

So,

→ Third term = 9

→ a•r^(3 - 1) = 9

→ a•r² = 9 -------- Equation (1)

and,

→ Fifth term = 16

→ a•r^(5 - 1) = 16

→ a•r⁴ = 16 -------- Equation (2)

dividing Equation (2) by Equation (1),

→ a•r⁴ ÷ a•r² = 16 ÷ 9

→ r⁴ - r² = 16 ÷ 9

→ r^(4 - 2) = 16 ÷ 9

→ r² = (16/9) ------- Equation (3)

→ r = ± (4/3)

putting value of Equation (3) in Equation (1),

→ a × (16/9) = 9

→ a = (81/16)

therefore taking r = (4/3) ,

→ Fourth term = a•r^(4 - 1)

→ a•r³

→ (81/16) × (4/3)³

→ (81/16) × (64/27)

→ 3 × 4

→ 12 (Ans.)

taking r = (-4/3)

→ a•r³

→ (81/16) × (-4/3)³

→ (81/16) × (-64/27)

→ 3 × (-4)

→ (-12) (Ans.)

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