To solve question 38, let's break it down:

The problem involves investing an initial amount and increasing it monthly, which suggests an arithmetic progression (AP).

Given:

• Initial installment = Rs. 6,000
• Monthly increase = Rs. 300
• Number of months = 24

We need to determine if:

1. The total value after the 13th installment is Rs. 119,900.
2. The value of the 13th installment is Rs. 9,900.

Step 1: Total Value after 13 Installments

The total amount received over the first 13 installments is the sum of the arithmetic progression (AP), where:

• First term $a = 6000$
• Common difference $d = 300$
• Number of terms $n = 13$

The sum $S_n$ of the first $n$ terms of an AP is given by: $S_n = \frac{n}{2} [2a + (n-1)d]$

Substituting the values: [ S_{13} = \frac{13}{2} [2(6000) + (13-1)(300)] = \frac{13}{2} [12000 + 3600] = \frac{13}{2} \times 15600 = 13 \times 7800 = 101,400 ]

Thus, the total after 13 installments is Rs. 101,400, not Rs. 119,900. Therefore, the first statement is false.

Step 2: Value of the 13th Installment

The value of any term in an AP is given by: $T_n = a + (n-1)d$ For the 13th installment: $T_{13} = 6000 + (13-1)(300) = 6000 + 12(300) = 6000 + 3600 = 9600$

Thus, the 13th installment is Rs. 9,600, not Rs. 9,900. Therefore, the second statement is false.

Conclusion:

Both statements are incorrect, so the correct answer is: d) Both statements are not correct.

Would you like more details or have any questions?
Here are some related questions to explore:

1. What is the total after 24 installments?
2. How does the common difference affect the total?
3. Can we calculate the value of the 24th installment?
4. How would the solution change if the increase was Rs. 500 instead of Rs. 300?
5. What if the installments decrease instead of increase?

Tip: For arithmetic progressions, always apply the sum formula carefully and verify individual terms with the general formula.