Question

4 consider the geometric series ( 7 + 14 + 28 + dots + 3584 ).
a find the number of terms in the series.
b hence find the sum of the series.
486 number sequences (chapter 22)

Explanation:

Part (a)

Step 1: Identify G.P. parameters

In a geometric progression (G.P.), the first term \( a = 7 \), common ratio \( r=\frac{14}{7} = 2 \), and the \( n \)-th term \( T_n=3584 \). The formula for the \( n \)-th term of a G.P. is \( T_n = ar^{n - 1} \).

Step 2: Substitute values into the formula

Substitute \( a = 7 \), \( r = 2 \), and \( T_n=3584 \) into \( T_n=ar^{n - 1} \):
\[
3584=7\times2^{n - 1}
\]

Step 3: Solve for \( n \)

First, divide both sides by 7:
\[
\frac{3584}{7}=2^{n - 1}
\]
\[
512 = 2^{n - 1}
\]
Since \( 512=2^9 \), we have:
\[
2^9=2^{n - 1}
\]
Equating the exponents (since the bases are the same):
\[
9=n - 1
\]
\[
n=10
\]

Part (b)

Step 1: Recall the sum formula for a G.P.

The sum of the first \( n \) terms of a G.P. is given by \( S_n=\frac{a(r^n - 1)}{r - 1} \) when \( r>1 \). Here, \( a = 7 \), \( r = 2 \), and \( n = 10 \).

Step 2: Substitute values into the formula

Substitute \( a = 7 \), \( r = 2 \), and \( n = 10 \) into \( S_n=\frac{a(r^n - 1)}{r - 1} \):
\[
S_{10}=\frac{7(2^{10}-1)}{2 - 1}
\]

Step 3: Simplify the expression

Since \( 2 - 1=1 \) and \( 2^{10}=1024 \):
\[
S_{10}=7\times(1024 - 1)
\]
\[
S_{10}=7\times1023
\]
\[
S_{10}=7161
\]

Answer:

s:
a) The number of terms \( n=\boxed{10} \)
b) The sum of the series \( S_{10}=\boxed{7161} \)