Alright, let’s break down and solve each part of the question step-by-step:
### Problem 1: Solving for [tex]\( t_n \)[/tex]
Given the sequence [tex]\( \{t_n\} \)[/tex] where [tex]\( S_n = n(3n + 2) \)[/tex], we need to find [tex]\( t_n \)[/tex].
1. [tex]\( S_n \)[/tex] represents the nth partial sum of the sequence [tex]\( t_n \)[/tex].
2. To find [tex]\( t_n \)[/tex], use the relation [tex]\( t_n = S_n - S_{n-1} \)[/tex].
Given [tex]\( S_n = n(3n + 2) \)[/tex], let’s express [tex]\( S_{n-1} \)[/tex]:
[tex]\[ S_{n-1} = (n-1)[3(n-1) + 2] = (n-1)(3n - 3 + 2) = (n-1)(3n - 1) \][/tex]
Now, subtract [tex]\( S_{n-1} \)[/tex] from [tex]\( S_n \)[/tex]:
[tex]\[ t_n = S_n - S_{n-1} \][/tex]
[tex]\[ t_n = n(3n + 2) - (n-1)(3n - 1) \][/tex]
[tex]\[ = 3n^2 + 2n - [3n^2 - 3n - n + 1] \][/tex]
[tex]\[ = 3n^2 + 2n - 3n^2 + 3n + n - 1 \][/tex]
[tex]\[ = 2n + 3n + n - 1 \][/tex]
[tex]\[ = 6n - 1 \][/tex]
Thus, [tex]\( t_n = 6n - 1 \)[/tex].
So, the correct answer is:
[tex]\[ \boxed{6n - 1} \][/tex]
### Problem 2: Yearly Income Calculation
Given:
- Initial yearly salary = Rs. 9600
- Annual increment = Rs. 40 per month, which is Rs. 480 per year (since every month increment is Rs. 40 and there are 12 months in a year).
- Duration of service = 30 years
To find the total amount received over the 30 years:
1. The salary in arithmetic progression where [tex]\( a = 9600 \)[/tex], [tex]\( d = 480 \)[/tex], and [tex]\( n = 30 \)[/tex].
Use the formula for the sum of an arithmetic series [tex]\( S_n = \frac{n}{2} [2a + (n-1)d] \)[/tex]:
[tex]\[ S_{30} = \frac{30}{2} [2(9600) + (30-1)(480)] \][/tex]
[tex]\[ = 15 [19200 + 29 \cdot 480] \][/tex]
[tex]\[ = 15 [19200 + 13920] \][/tex]
[tex]\[ = 15 \cdot 33120 \][/tex]
[tex]\[ = 496800 \][/tex]
So, the total amount he will receive in his service of 30 years is:
[tex]\[ \boxed{496800} \][/tex]
### Problem 3: Nitin's Repayment Analysis
Nitin borrowed Rs. 1000 and repays in daily installments over a month.
- Initial repayment: Rs. 5
- Daily increase: Rs. 2
- Days in the month: 30
The sequence of daily payments forms an arithmetic series where [tex]\( a = 5 \)[/tex], [tex]\( d = 2 \)[/tex], and [tex]\( n = 30 \)[/tex].
To find the total amount repaid over 30 days, use the sum of an arithmetic series formula again:
[tex]\[ S_{30} = \frac{30}{2} [2(5) + (30-1)2] \][/tex]
[tex]\[ = 15 [10 + 58] \][/tex]
[tex]\[ = 15 \cdot 68 \][/tex]
[tex]\[ = 1020 \][/tex]
Now, calculate the net gain or loss:
[tex]\[ \text{Net gain/loss} = \text{Total paid} - \text{Initial loan} \][/tex]
[tex]\[ = 1020 - 1000 \][/tex]
[tex]\[ = 20 \][/tex]
Since the net amount is positive, it's a gain. Therefore, the correct answer is:
[tex]\[ \boxed{\text{Gain Rs. 20}} \][/tex]
