Previous year questions: Sequences and Series

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Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the source and the marks. The stems were checked against the official PDF pages and every value against the official key. Before computing, decide whether the progression is arithmetic or geometric, then write its first term and common difference or ratio.

Q1 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Between 59 and 159 there are 39 arithmetic means A1, A2, ..., A39. Find the mean of A25, A28, A31 and A36.

Answer: The 41 terms from 59 to 159 form an arithmetic progression, so the common difference is (159 minus 59) by 40 equal to 2.5. The kth inserted mean is 59 plus 2.5k. Then A25 is 59 plus 62.5 equal to 121.5, A28 is 59 plus 70 equal to 129, A31 is 59 plus 77.5 equal to 136.5, and A36 is 59 plus 90 equal to 149. Their average is (121.5 plus 129 plus 136.5 plus 149) by 4, which is 536 by 4 equal to 134. The official answer is 134.

Q2 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

Find the value of 1 cubed minus 2 cubed plus 3 cubed minus 4 cubed and so on up to 15 cubed.

Answer: Separate the odd and even cubes. The sum of the cubes of the first 15 natural numbers is the square of 15 times 16 by 2, that is 120 squared equal to 14400. The even cubes are 8 times the cubes of 1 to 7, and the sum of the first seven cubes is 28 squared equal to 784, so the even contribution is 6272. The alternating sum is 14400 minus twice 6272, which is 14400 minus 12544 equal to 1856. The official key is 1856.

Q3 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

The sum of the first ten terms of an arithmetic progression is 160, and the sum of the first two terms of a geometric progression is 8. The first term of the arithmetic progression equals the common ratio of the geometric progression, and the first term of the geometric progression equals the common difference of the arithmetic progression. Find the sum of all possible values of the first term of the geometric progression.

Answer: For the arithmetic progression, 5 times (2a plus 9d) equals 160, so 2a plus 9d equals 32. For the geometric progression, b plus ba equals 8, that is b(1 plus a) equals 8, and b equals d. Substituting d equal to 8 by (1 plus a) gives 2a plus 72 by (1 plus a) equals 32. Multiplying by (1 plus a) gives 2a squared minus 30a plus 40 equal to 0, that is a squared minus 15a plus 20 equal to 0. The two roots satisfy a1 plus a2 equal to 15 and a1a2 equal to 20. The corresponding b values are 8 by (1 plus a1) and 8 by (1 plus a2), and their sum is 8 times (2 plus a1 plus a2) divided by (1 plus a1)(1 plus a2), which is 8 times 17 divided by 36, equal to 34 by 9. The official key is 34 by 9.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Let alpha and beta be the roots of x squared minus x plus p equal to 0, and gamma and delta the roots of x squared minus 4x plus q equal to 0, where p and q are integers. If alpha, beta, gamma and delta are in geometric progression in that order, find the absolute value of p plus q.

Answer: Write the four terms as a, ar, ar squared and ar cubed. The first pair sums to 1 and the second pair sums to 4, so a(1 plus r) equals 1 and ar squared (1 plus r) equals 4. Dividing gives r squared equal to 4, so r is 2 or minus 2. With r equal to 2 the first term is 1 by 3, and p is a squared r equal to 2 by 9, which is not an integer, so this case is rejected. With r equal to minus 2 the first term is minus 1, p is a squared r equal to minus 2, and q is a squared times r to the fifth equal to minus 32. Then p plus q is minus 34 and its absolute value is 34. The official key is 34.

Q5 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

The sum of the first n terms of an arithmetic progression is 3n squared plus 5n. Find the sum of the squares of the first ten terms.

Answer: The nth term is the difference of consecutive sums: 3n squared plus 5n minus 3(n minus 1) squared minus 5(n minus 1), which simplifies to 6n plus 2. The first ten terms are 8, 14, 20 and so on up to 62. The sum of their squares is the sum of (6n plus 2) squared for n from 1 to 10, that is 36 times the sum of the first ten squares plus 24 times the sum of the first ten numbers plus 40. Using 385 for the sum of squares and 55 for the sum, the total is 13860 plus 1320 plus 40 equal to 15220. The official key is 15220.

Q6 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

Find the sum 1 cubed by 1 plus (1 cubed plus 2 cubed) by (1 plus 3 plus 5) plus (1 cubed plus 2 cubed plus 3 cubed) by (1 plus 3 plus 5 plus 7) and so on up to 8 terms.

Answer: The numerator of the nth term is the sum of the first n cubes, which equals n squared (n plus 1) squared by 4. The denominator is the sum of the first n odd numbers, which equals n squared. So the nth term is (n plus 1) squared by 4. Summing from n equal to 1 to 8 gives one quarter of (2 squared plus 3 squared up to 9 squared). The sum of squares from 1 to 9 is 285, and subtracting the 1 squared term gives 284, so the total is 284 by 4 equal to 71. The official key is 71.

Q7 (NCERT Exemplar, 3 marks)

A man saved 66000 rupees in 20 years. In each year after the first he saved 200 rupees more than in the previous year. How much did he save in the first year?

Answer: The yearly savings form an arithmetic progression with first term a and common difference 200. The sum of 20 terms is 10 times (2a plus 19 times 200), equal to 66000. So 2a plus 3800 equals 6600, giving 2a equal to 2800 and a equal to 1400. He saved 1400 rupees in the first year. Source: NCERT Exemplar Class 11 Mathematics, Chapter 9, Exercise 9.3, Short Answer Type, Q2.

Q8 (NCERT Exemplar, 3 marks)

A man accepts a position with an initial salary of 5200 rupees per month, with an automatic increase of 320 rupees in the very next month and each month after. Find his salary for the tenth month and his total earnings in the first year.

Answer: The monthly salaries form an arithmetic progression with first term 5200 and common difference 320. The tenth month salary is 5200 plus 9 times 320 equal to 5200 plus 2880, that is 8080 rupees. The first year has 12 months, and their sum is 6 times (2 times 5200 plus 11 times 320), which is 6 times (10400 plus 3520) equal to 6 times 13920, that is 83520 rupees. Source: NCERT Exemplar Class 11 Mathematics, Chapter 9, Exercise 9.3, Short Answer Type, Q3.

Q9 (NCERT Exemplar, 3 marks)

A carpenter was hired to build 192 window frames. On the first day he made five frames, and each day after he made two more frames than the day before. How many days did it take him to finish the job?

Answer: The daily output forms an arithmetic progression with first term 5 and common difference 2. If the job takes n days, the sum is n by 2 times (10 plus 2(n minus 1)), equal to 192. Multiplying out gives 2n squared plus 8n minus 384 equal to 0, or n squared plus 4n minus 192 equal to 0. The positive root is 12, since 12 squared plus 4 times 12 is 192. He took 12 days. Source: NCERT Exemplar Class 11 Mathematics, Chapter 9, Exercise 9.3, Short Answer Type, Q5.

Q10 (NCERT Exemplar, 3 marks)

An equilateral triangle has side 20 cm. A second equilateral triangle is inscribed by joining the midpoints of its sides, and the process is repeated. Find the perimeter of the sixth inscribed equilateral triangle.

Answer: Each new triangle has side half of the previous one, so its perimeter is also half of the previous perimeter. The first perimeter is 60 cm, and the perimeters form a geometric progression with ratio 1 by 2. The sixth perimeter is 60 times 1 by 2 to the fifth, that is 60 by 32 equal to 15 by 8 cm. Source: NCERT Exemplar Class 11 Mathematics, Chapter 9, Exercise 9.3, Short Answer Type, Q7.