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.

The fourth term of an A.P. is equal to three times the first term and the seventh term exceeds twice the third term by 1. Find the first term and the common difference of the A.P.

Fourth term = a + 3d So, a + 3d = 3a 2a = 3d, a = (3/2)d ------- (1) Seventh term = a + 6d Third term = a + 2d So, a + 6d ─ 2(a + 2d) = 1 a +6d ─ 2a ─ 4d =1 ─ a + 2d =1 Substituting value of a from Eq. 1, ─ (3/2)d + 2d = 1 ─ 3d + 4d = 2 i.e., d = 2 Substituting the value of d in eq.1, we get a = 3 So the correct option is Option 3.

In a certain A.P., the 24th term is twice the 10th term. Is it true that the 72nd term is twice the 34th term?

24 th term = a + 23d 10 th term = a + 9d So, a + 23d = 2(a+9d) a + 23d = 2a + 18d -------- (1) Now, 72nd term = a + 71d To get this, adding 48 d to both sides of eq. 1 So, a + 23d + 48d = 2a + 18d + 48d a + 71d = 2a + 66d= 2(a + 33d) so, it is true that the 72nd term is twice the 34th term. So the correct option is Option 2.