Question

The sum of the series 1/2+ 1 + 2 + .... to 10 terms is


511.5


1024


512


1023.5

Answer 1

Proper question.

Find the sum of the first 10 terms where the geometric sequence $\{a_{n}\}$ is given by $a=\dfrac{1}{2}$(first term) and $r=2$(common ratio).

Solution.

Since we have the first term $a=\dfrac{1}{2}$ and the common ratio $r=2$, the series is $S_{n}=\dfrac{\dfrac{1}{2} (2^{n}-1)}{2-1} =\dfrac{1}{2} (2^{n}-1)$.

The sum of the first 10 terms is $S_{10}=\dfrac{1}{2} (2^{10}-1)=2^{9}-\dfrac{1}{2} =\boxed{511.5}$.

Formula.

Let the series of the geometric series be $S_{n}=a+ar+...+ar^{n-1}$. Then, $rS_{n}=ar+ar^2+...+ar^{n}$. Subtracting both equations we derive $(r-1)S_{n}= ar^{n}-a$. The series of the geometric series is $S_{n}=\dfrac{a(r^{n}-1)}{r-1}$(where $r\neq 1$).

Short note.

Sequence: Numbers arranged in a certain order.

Series: Sum of the sequence.

Answer 2

Given : A.P ½ , 1 , 3/2 . . . .

• First Term = a = ½
• Common Diiference = d = 1 - ½ = ½
• n = 10 terms

$\huge \sf { \: }^{ \huge a} { \large \: n}^{ \: } = a + (n - 1)d$

$\large\sf { \: }^{ \huge \: a} {10}^{ \: } = (\frac{1}{2}) + (10 - 1)(0.5) =$

$\sf = \: 0.5 + 9(0.5) = 0.5 + 4.5 = 5$