Sequences and Series

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Sequences and Series — solved practice questions

8 JEE Mathematics questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   The sum of the first 20 terms of the arithmetic progression 2, 5, 8, 11, ... is:

   • A590
   • B600
   • C610
   • D620
   Show answer & solution

   Correct answer: (C) 610

   Here a = 2, d = 3, n = 20. Sum = (n/2)[2a + (n-1)d] = 10[4 + 57] = 610.

2. Q2easy

   The geometric mean of the numbers 4 and 9 is:

   • A6
   • B6.5
   • C5
   • D36
   Show answer & solution

   Correct answer: (A) 6

   The geometric mean of two positive numbers is sqrt(4 · 9) = sqrt(36) = 6. (The value 6.5 is the arithmetic mean, not the geometric mean.)

3. Q3easy

   The sum of the cubes 1^3 + 2^3 + 3^3 + 4^3 + 5^3 is:

   • A125
   • B200
   • C215
   • D225
   Show answer & solution

   Correct answer: (D) 225

   Sum of cubes of first n natural numbers = (n(n+1)/2)^2. For n = 5 this is (15)^2 = 225.

4. Q4medium

   In a geometric progression the 3rd term is 12 and the 6th term is 96. The sum of the first 8 terms is:

   • A510
   • B765
   • C768
   • D1020
   Show answer & solution

   Correct answer: (B) 765

   Dividing the terms gives r^3 = 96/12 = 8, so r = 2, and a·r^2 = 12 gives a = 3. Sum = 3(2^8 - 1)/(2 - 1) = 3·255 = 765.

5. Q5medium

   The sum of the infinite geometric series 4 + 4/3 + 4/9 + 4/27 + ... is:

   • A4
   • B5
   • C6
   • D12
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   Correct answer: (C) 6

   First term a = 4 and common ratio r = 1/3, so the sum to infinity is a/(1 - r) = 4/(2/3) = 6.

6. Q6medium

   The sum of the series 1/(1·2) + 1/(2·3) + 1/(3·4) + ... + 1/(10·11) is:

   • A10/11
   • B9/10
   • C11/12
   • D1
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   Correct answer: (A) 10/11

   Each term 1/(k(k+1)) = 1/k - 1/(k+1) telescopes, so the sum is 1 - 1/11 = 10/11.

7. Q7medium

   The 7th term of an arithmetic progression is 34 and the 13th term is 64. Its 18th term is:

   • A79
   • B84
   • C94
   • D89
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   Correct answer: (D) 89

   From a + 6d = 34 and a + 12d = 64, we get 6d = 30 so d = 5 and a = 4. The 18th term = a + 17d = 4 + 85 = 89.

8. Q8medium

   The number of terms in the arithmetic progression 3, 7, 11, ..., 407 is:

   • A100
   • B102
   • C101
   • D104
   Show answer & solution

   Correct answer: (B) 102

   With a = 3 and d = 4, the nth term is 3 + (n-1)·4 = 407, giving n - 1 = 101, so n = 102.

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