Let's break down the problem step by step to solve each part:
### Part (a) - Identifying the Geometric Sequence
We need to determine whose expenditure (Anshu's or Shreeanshu's) follows a geometric sequence.
- Geometric Sequence: In a geometric sequence, each term is found by multiplying the previous term by a constant ratio [tex]\( r \)[/tex].
Expenditure of Anshu:
- First day: Rs 150
- Second day: Rs 180
- Third day: Rs 210
To check if this is a geometric sequence, we calculate the ratio between consecutive terms:
[tex]\[
\text{Ratio between first and second day} = \frac{180}{150} = 1.2
\][/tex]
[tex]\[
\text{Ratio between second and third day} = \frac{210}{180} = 1.167
\][/tex]
Since the ratios are not equal, Anshu's expenditure does not follow a geometric sequence.
Expenditure of Shreeanshu:
- First day: Rs 1
- Second day: Rs 2
- Third day: Rs 4
To check if this is a geometric sequence, we calculate the ratio between consecutive terms:
[tex]\[
\text{Ratio between first and second day} = \frac{2}{1} = 2
\][/tex]
[tex]\[
\text{Ratio between second and third day} = \frac{4}{2} = 2
\][/tex]
Since the ratios are the same, Shreeanshu's expenditure follows a geometric sequence with a common ratio [tex]\( r = 2 \)[/tex].
Answer for part (a):
Shreeanshu
### Part (b) - Total Expenditure of Anshu up to 12 days
Since Anshu's expenditure increases by a fixed amount each day (not a geometric progression), it follows an arithmetic sequence.
- Arithmetic Sequence: In an arithmetic sequence, each term is found by adding a constant difference [tex]\( d \)[/tex] to the previous term.
Expenditure of Anshu:
- First day: Rs 150
- Second day: Rs 180
- Third day: Rs 210
Calculate the common difference [tex]\( d \)[/tex]:
[tex]\[
d = 180 - 150 = 30
\][/tex]
[tex]\[
d = 210 - 180 = 30
\][/tex]
Since the common difference [tex]\( d \)[/tex] is constant, it's an arithmetic sequence with:
- First term [tex]\( a = 150 \)[/tex]
- Common difference [tex]\( d = 30 \)[/tex]
The formula for the sum [tex]\( S \)[/tex] of the first [tex]\( n \)[/tex] terms in an arithmetic sequence is:
[tex]\[
S_n = \frac{n}{2} \left(2a + (n-1)d\right)
\][/tex]
For [tex]\( n = 12 \)[/tex] days:
[tex]\[
S_{12} = \frac{12}{2} \left(2 \cdot 150 + (12-1) \cdot 30\right)
\][/tex]
[tex]\[
S_{12} = 6 \left(300 + 330\right)
\][/tex]
[tex]\[
S_{12} = 6 \cdot 630 = 3780
\][/tex]
Answer for part (b):
Rs 3780
In summary:
a) Shreeanshu's expenditure is in a geometric sequence.
b) The total expenditure of Anshu up to 12 days is Rs 3780.
