Question

5. How many consecutive terms starting from the first term of the series
(i) 3 +9+ 27 +...... would sum to 1092 ? (ii) 2 +6+ 18 + .... would sum to 728 ?

Answer 1

(i) 3 + 9 + 27 + ...... would sum to 1092
here series is 3 + 3² + 3³ + .......n terms = 1092
Clearly shown it is in GP . Where first term, a = 3 and common ratio ,r = 3
Now, Sn = a[rⁿ - 1]/(r - 1)
1092 = 3[3ⁿ - 1]/(3 - 2)
1092 = 3[3ⁿ - 1]/2
2184/3 = 3ⁿ - 1
728 + 1 = 3ⁿ
729 = 3⁶ = 3ⁿ
n = 6

(ii) 2 + 6 + 18 + ..... would sum to 728
here series 2 + 2(3) + 2(3)² + 2(3)³ + ...... = 728
It is also in GP where first term , a = 2 and common ratio, r = 3
Now, Sn = a[rⁿ - 1]/(r - 1)
728 = 2[3ⁿ - 1]/(3 - 2)
728 = 3ⁿ - 1
729 = 3⁶ = 3ⁿ
n = 6

Answer 2

$\mathfrak{HELLO\:\: DEAR,}$

$\boxed{1}.$ 3 + 9 + 27 + ......... would sum to 1092?

here the the progression is in gp.
so, a = 3 , r = 9/3 = 3 , $\mathbf{S_n = 1092}$

$\mathbf{S_n = \frac{a(r^n - 1)}{r - 1}}$

$\mathbf{1092 = \frac{3(3^n - 1)}{3 - 1}}$

$\mathbf{1092 = 3/2 * (3^n - 1)}$

$\mathbf{1092*2/3 = (3^n - 1)}$

$\mathbf{ 728 = 3^n - 1}$

$\mathbf{728 + 1= 3^n}$

$\mathbf{3^6 = 3^n}$

$\mathbf{ n = 6}$

$\boxed{2}.$ 2 +6+ 18 + .... would sum to 728 ?

here, the the progression is in gp.
so, a = 2 , r = 6/2 = 3 , $\mathbf{S_n = 728}$

$\mathbf{S_n = \frac{a(r^n - 1)}{r - 1}}$

$\mathbf{728 = \frac{2(3^n - 1)}{3 - 1}}$

$\mathbf{728 + 1 = 3^n}$

$\mathbf{3^6 = 3^n}$

$\mathbf{n = 6}$

I HOPE ITS HELP YOU DEAR,
THANKS