Answer 1

3 + 6 + 12 + …. + 1536

a = 3, r = `6/3` = 2

t_n = 1536

a.r^n–1 = 1536 ⇒ 3(2^n–1) = 1536

2^n–1 = `1536/3` ⇒ 2^n–1 = 512

2^n–1 =  2⁹

∴ n – 1 = 9

n = 9 + 1 = 10

Number of terms = 10

S_n = `("a"("r"^"n" - 1))/("r" - 1)`

S_n = `(3(2^10 - 1))/(2 - 1)`

= `(3(1024 - 1))/1`

= 3 × 1023

= 3069

∴ Sum of the series is 3069