Question

Find the least value of n for which 1 + 3 + 9 + 27 + ….. to n terms exceeds 1000.

Answer 1

Given series is GP
First term, a = 1
Common ratio, r = 3
Sum to n terms is
$\frac{a( {r}^{n} - 1)}{r - 1} \: \: should \: be \: \: > 1000 \\ \frac{1 \times ( {3}^{n} - 1)}{3 - 1} > 1000 \\ {3}^{n} - 1 > 2000 \\ {3}^{n} > 2001 \\ n > 6$
Therefore least value of n is 7

Answer 2

The least value of "n is equal to 7".

Step-by-step explanation:

The sum of the sequence are:

1 + 3 + 9 + 27 + ….. to n

$S_{n} =1000$

To find, the least value of n = ?

The given sequence are in GP.

Here, first term(a) = 1 ,  common ratio(r) = $\dfrac{3}{1} =3$

We know that,

The sum of nth term of GP(r > 1),

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

⇒ $\dfrac{1(3^{n}-1)}{3-1}>1000$

⇒ $3^{n}-1>2\times 1000$

⇒ $3^{n}>2000+1>2001$]

[ ∵$3^{7} =2187$]

The least value of n = 7,  satisfied this inequalities.

Hence, the least value of "n is equal to 7".