What is the number of terms in the A.P. given below? 2, 5, 8, ..., 59

2, 5, 8, …, 59 a = 2, d = 5 ─ 2 = 3, last term = 59 aₙ = a + (n ─ 1)d 59 = 2 + (n ─ 1) 3 59 ─ 2 = 3n ─ 3 57 + 3 = 3n 60 = 3n 20 = n i.e., n = 20 So the correct answer is Option 3 If option 1 is chosen then it is incorrect. The student did not change the sign of 3 while transposing it to LHS If Option 2 is chosen then also it is incorrect. The student must have took 59 as n and did the calculation for nth term If Option 4 is chosen then also it is incorrect. The student must have interchanged values of a and d in formula

Which term of the A.P. 1/3, 5/3, 9/3,13/3, … is 37/3?

1/3, 5/3, 9/3, 13/3, … is an infinite A.P. For this AP, a = 1/3, d = 4/3 We know that, aₙ = a + (n ─ 1)d So, 37/3 = a + (n ─ 1) d 37/3 = 1/3 + (n ─ 1)4 / 3 Multiplying throughout with 3, 37 = 1 + (n ─ 1)4 36 = 4n ─ 4 40 = 4n n = 10 so the correct option is Option 1 If you have chosen Option 2, then it is incorrect. You must have forgotten to change the sign of ─ 4 while transposing it to LHS If you have chosen Option 3, then it is also incorrect. It is possible that the values of a and d were interchanged in the formula.

The 11th and 13th terms of an AP are 35 and 41 respectively. The common difference of the A.P. is

11th term is --- a + (11 ─ 1)d = 35 a + 10d = 35 a = 35 ─ 10d ……… (1) 13th term is -----a + (13 ─ 1)d = 35 a + 12d = 41 ………… (2) Substituting value of a from eq 1 to eq2, 35 ─ 10d + 12d = 41 2d = 41 ─ 35 .....(3) 2d = 6..... (4) d = 3 So option 4 is the right answer If Option 1 is chosen, it is incorrect. It is possible that the student did not change sign of 35 while transposing it to RHS to get eq. 3 If Option 2 is chosen, then also it is incorrect. The student must have determined the values of d and a but reported result for a If Option 3 is chosen, then also it is incorrect. It is possible that the student did not divide eq. 4 by 2 to get the correct value of d.

How many three digit numbers are divisible by 5?

Three digit numbers divisible by 5 can be given by the list 100, 105, 110, ..., 995, where 995 is the biggest three digit number divisible by 5. This is an AP with a = 100, d = 5, last term = 995 By using the formula, aₙ = a + (n ─ 1)d, we get, 995 = 100 + (n ─ 1)5 895 = 5n ─ 5 5n = 900 n = 180 So Option 2 is the right answer. If option 1 is chosen then it is incorrect. Here the last term would have been taken as 1000 which is incorrect because here we are talking about only 3-digit numbers divisible by 5 If Option 3 is chosen then also it is incorrect. Here the sign of ─5 is not changed while transposing it to LHS

Kedar's height was 70cm when he was one year old. His height increased to 80cm at 2 years, 90 cm at 3 years and he was 100 cm when he was 4 years old. If the increase in height happens at the same rate, what will be his height when he is 15 years old?

We can list Kedar's heights in a sequence as 70, 80,90,100, … This is an AP with a = 70 , d = 10 When Kedar will be 15 years old, n = 15 a₁₅ = 70 + ( 15 ─ 1)10 = 70 + 14(10) a₁₅ = 70 + 140 = 210 cm So the correct option is Option 3 If Option 1 is chosen then it is incorrect. Here a and d were interchanged in the formula If Option 2 is chosen then also it is incorrect. Here if n was taken as 10 instead of 15.