Question

The 11^th term of the geometric series \(\displaystyle\sum^{20}_{r=0}2\times(-2)^r\) is equal to

• A. -4096
• B. 1024
• C. 2048
• D. 1048
• E. -2024

Answer 1

The Correct Option is C

The general form of the $n$\textsuperscript{th} term in a geometric series is given by: \[ T_n = a \cdot r^{n-1} \] where $a$ is the first term and $r$ is the common ratio. Here, the series is given by: \[ 2 \times (-2)^r \] Thus, the first term $a = 2 \times (-2)^0 = 2$, and the common ratio $r = -2$. We are asked to find the 11\textsuperscript{th} term, so we apply the formula for the $n$\textsuperscript{th} term: \[ T_{11} = 2 \cdot (-2)^{10} \] Now, calculate $(-2)^{10}$: \[ (-2)^{10} = 1024 \] Thus, the 11\textsuperscript{th} term is: \[ T_{11} = 2 \cdot 1024 = 2048 \]

The correct option is (C) : \(2048\)

Answer 2

The given series is \(\displaystyle\sum^{20}_{r=0}2\times(-2)^r\). This is a geometric series with the general term \(a_r = 2 \times (-2)^r\).

The first term (r=0) is \(a_0 = 2 \times (-2)^0 = 2 \times 1 = 2\).

The second term (r=1) is \(a_1 = 2 \times (-2)^1 = 2 \times (-2) = -4\).

The common ratio (r) is \(\frac{a_1}{a_0} = \frac{-4}{2} = -2\).

We want to find the 11^th term of this geometric series. Note that the index starts at r=0, so the 11^th term corresponds to r = 10.

The general formula for the nth term (starting from n=1) of a geometric series is \(a_n = a_1 \times r^{n-1}\). However, since our index starts from r=0, the (r+1)-th term of our series is given by:

\(a_r = 2 \times (-2)^r\)

To find the 11^th term, we have \(r = 10\) :

\(a_{10} = 2 \times (-2)^{10} = 2 \times 1024 = 2048\)

Therefore, the 11^th term of the geometric series is equal to 2048.