The third and the sixth term of a geometric progression are 24 and 64/9 respectively. Find a) the first term and common ratio
b) the sum of the first five terms

Ai. Common ratio = 2/3

Aii. First term = 54

B. Sum of the first five terms = 422/3

Step-by-step explanation:

From the question given above, the following data were obtained:

3rd term (T3) = 24

6Th term (T6) = 64/9

First term (a) =?

Common ratio (r) =?

Sum of the first five terms (S5) =?

Ai. Determination of the common ratio (r).

T3 = ar²

T3 = 24

24 = ar² (1)

T6 = ar⁵

T6 = 64/9

64/9 = ar⁵ (2)

The equation are:

24 = ar² (1)

64/9 = ar⁵ (2)

Divide equation 2 by equation 1.

64/9 ÷ 24 = ar⁵ / ar²

64/9 × 1/24 = r³

8/27 = r³

Take the cube root of both side

r = 3√(8/27)

r = 2/3

Thus, the common ratio is 2/3

Aii. Determination of the first term (a).

T3 = ar²

3rd term (T3) = 24

Common ratio (r) = 2/3

First term (a) =?

24 = a(2/3)²

24 = 4a/9

Cross multiply

24 × 9 = 4a

216 = 4a

Divide both side by 4

a = 216/4

a = 54

Thus, the first term (a) is 54

B. Determination of the sum of the first five terms.

Common ratio (r) = 2/3

First term (a) = 54

Number of term (n) = 5

Sum of first five terms (S5) =?

Sn = a[1 –rⁿ] / 1 – r

S5 = 54[1 – (⅔)⁵] / 1 – ⅔

S5 = 54 [1 – 32/243] / ⅓

S5 = 54 (211/243) × 3

S5 = 54 × 211/81

S5 = 6 × 211/9

S5 = 2 × 211/3

S5 = 422/3

Thus, the sum of the first five terms is 422/3