Question

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of the first 10 terms in the sequence, then T isA. Greater than 2B. Between 1 and 2C. Between 1/2 and 1D. Between 1/4 and 1/2E. Less than 1/4

Answer 1

Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is $(-1)^{k+1}(\frac{1}{2^{k}})$.

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as $(-1)^{k+1}(\frac{1}{2^{k}})$.

where k is from 1 to 10.

By the given expression sequence will be $\frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......$

In this sequence first term "a" = $\frac{1}{2}$

and common ratio in each successive term to the previous term is 'r' = $\frac{\frac{(-1)}{4}}{\frac{1}{2} }$

r = $-\frac{1}{2}$

Since the sequence is infinite and the formula to calculate the sum is represented by

$S=\frac{a}{1-r}$ [Here r is less than 1]

$S=\frac{\frac{1}{2} }{1+\frac{1}{2}}$

$S=\frac{\frac{1}{2}}{\frac{3}{2} }$

S = $\frac{1}{3}$

Now we are sure that the sum of infinite terms is $\frac{1}{3}$.

Therefore, sum of 10 terms will not exceed $\frac{1}{3}$

Now sum of first two terms = $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Now we are sure that sum of first 10 terms lie between $\frac{1}{4}$ and $\frac{1}{3}$

Since $\frac{1}{2}>\frac{1}{3}$

Therefore, Sum of first 10 terms will lie between $\frac{1}{4}$ and $\frac{1}{2}$.

Option D will be the answer.