Word problems based on APs

Correct answer: C. 205

The daily numbers of instruments form an arithmetic progression: 7, 10, 13, \ldots. Here, the first term is \(a=7\), the common difference is \(d=3\), and the number of terms is \(n=10\). Therefore, \(S_{10}=\frac{10}{2}[2(7)+(10-1)(3)]=5(14+27)=205\). Hence, 205 instruments are placed in 10 days. An answer such as 200 usually results from using an incorrect number of intervals or an incorrect value of the common difference. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.

Correct answer: B. 50 GB

The daily data amounts form an arithmetic progression because the amount increases by the same quantity, 3 GB, each day. Here the first term is a = 5 and the common difference is d = 3. To find the amount on the 16th day, use the nth-term formula a_n = a + (n - 1)d. Substituting the values gives a_16 = 5 + (16 - 1) × 3 = 5 + 15 × 3 = 5 + 45 = 50 GB. Hence option B is correct. Option A results from using only 14 increases, while option C adds one extra 3-GB increase; option D reflects two extra increases. The key is that the first day is the first term, so only 15 increases occur before day 16.

Correct answer: B. 12

The governing concept is the nth term of an arithmetic progression, because the fee increases by the same amount, ₹75, every month. The first term is a = 250 and the common difference is d = 75. For the fee in the nth month, use a_n = a + (n − 1)d. Substituting the required fee gives 1075 = 250 + 75(n − 1). Therefore, 825 = 75(n − 1), so n − 1 = 11 and n = 12. Hence the fee becomes ₹1075 in the twelfth month, making option B correct. The sum formula is not needed because the question asks for the fee in one particular month, not the total fee collected over several months. The other options would correspond to using an incorrect number of monthly increases.