Questions: Sequences and Series
Sequences and Series24 min readLesson 5 of 6
Q1. Write the first five terms of the sequence whose nth term is 2n + 1.
Answer: Put n equal to 1, 2, 3, 4 and 5. The terms are 3, 5, 7, 9 and 11. This is an arithmetic progression with first term 3 and common difference 2, so the same terms could have been produced by adding 2 repeatedly to 3 instead of substituting into the formula.
Hinglish: n = 1, 2, 3, 4, 5 rakho: 3, 5, 7, 9, 11. Ye AP hai jiska first term 3 aur common difference 2 hai.
Q2. Find the 10th term of the geometric progression 2, 6, 18 and so on.
Answer: The first term is 2 and the ratio is 3, since 6 by 2 is 3. The tenth term is 2 times 3 to the ninth, that is 2 times 19683, which equals 39366.
Hinglish: First term 2, ratio 3. 10th term = 2 × 3^9 = 2 × 19683 = 39366.
Q3. Find the sum of the first 8 terms of the geometric progression 1, 2, 4 and so on.
Answer: The first term is 1, the ratio is 2 and n is 8. The sum is 1 times (2 to the eighth minus 1) divided by (2 minus 1), which is 256 minus 1 equal to 255.
Hinglish: a = 1, r = 2, n = 8. Sum = (2^8 − 1)/(2 − 1) = 255.
Q4. Which term of the geometric progression 3, 6, 12 and so on equals 768.
Answer: With first term 3 and ratio 2, the nth term is 3 times 2 to the (n minus 1). Setting this equal to 768 gives 2 to the (n minus 1) equal to 256, so n minus 1 is 8 and n is 9. The ninth term is 768.
Hinglish: 3 × 2^(n−1) = 768 se 2^(n−1) = 256, yaani n − 1 = 8 aur n = 9. Nauvan term 768 hai.
Q5. Three numbers are in geometric progression with product 27 and sum 13. Find them.
Answer: Write them as a by r, a and ar. The product is a cubed equal to 27, so a is 3. The sum gives 3 times (1 by r plus 1 plus r) equal to 13, that is r plus 1 plus 1 by r equal to 13 by 3. Multiplying by 3r gives 3r squared minus 10r plus 3 equal to 0, whose roots are 3 and 1 by 3. The numbers are 1, 3 and 9 in either order.
Hinglish: Numbers a/r, a, ar maano. Product a³ = 27 se a = 3. Sum se r + 1 + 1/r = 13/3, quadratic 3r² − 10r + 3 = 0, roots 3 aur 1/3. Numbers 1, 3, 9.
Q6. Insert one geometric mean between 4 and 25.
Answer: The geometric mean is the square root of the product, which is the square root of 100, that is 10. The three numbers 4, 10 and 25 are in geometric progression with ratio 5 by 2, since 10 by 4 and 25 by 10 both equal 5 by 2. If the two ratios disagreed, the inserted number would not be the geometric mean.
Hinglish: GM = root of (4 × 25) = 10. 4, 10, 25 GP me hain, ratio 5/2.
Q7. If the arithmetic mean of two positive numbers is 10 and the geometric mean is 8, find the numbers.
Answer: The numbers have sum 20 and product 64, so they are the roots of x squared minus 20x plus 64 equal to 0. The roots are 4 and 16. Both are positive and their geometric mean is the square root of 64 equal to 8, which matches.
Hinglish: Sum 20 aur product 64 se x² − 20x + 64 = 0, roots 4 aur 16. GM root 64 = 8 sahi hai.
Q8. Find the sum of the first 6 terms of the geometric progression 5, 5 by 2, 5 by 4 and so on.
Answer: The first term is 5 and the ratio is 1 by 2. The sum is 5 times (1 minus 1 by 2 to the sixth) divided by (1 minus 1 by 2), which is 5 times (1 minus 1 by 64) divided by 1 by 2, that is 10 times 63 by 64 equal to 315 by 32.
Hinglish: a = 5, r = 1/2. Sum = 5(1 − 1/64)/(1/2) = 10 × 63/64 = 315/32.
Q9. A machine costing 15625 depreciates by 20 percent each year. Find its value after 5 years.
Answer: Each year the value is multiplied by 0.8, so after 5 years the value is 15625 times 0.8 to the fifth. Since 0.8 is 4 by 5, this is 15625 times 1024 by 3125, which equals 5120. The yearly multipliers form a geometric progression, which is why the fifth power appears instead of a simple subtraction.
Hinglish: Har saal value 0.8 se multiply hoti hai. 5 saal baad = 15625 × (0.8)^5 = 5120.
Q10. The sum of the first n terms of a sequence is given by 3n squared plus 5n. Find its nth term.
Answer: The nth term is the sum of the first n terms minus the sum of the first n minus 1 terms. Substituting gives 3n squared plus 5n minus 3(n minus 1) squared minus 5(n minus 1). Expanding and cancelling leaves 6n plus 2, so the sequence is 8, 14, 20 and so on, an arithmetic progression. A useful check is to recompute the sum from this term: the sum of 6k plus 2 for k from 1 to n is 3n squared plus 5n, which matches the given formula.
Hinglish: a_n = S_n − S_(n−1) = 3n² + 5n − [3(n−1)² + 5(n−1)] = 6n + 2. Sequence 8, 14, 20, ...
Common mistakes on this chapter
Using the geometric sum formula with r equal to 1 is the first error, because the denominator vanishes. Confusing a term with a sum is the second. Writing three terms in geometric progression as a, ar and ar squared is the third; it works, but the product becomes a cubed times r cubed and the symmetry is lost, which often turns an easy question into a hard one. Forgetting that fixed-percentage growth is geometric rather than arithmetic is the fourth. Read the answer against these four checks, and keep the working for the sum formula next to the formula itself so a slip is easy to spot.
Quiz
Quiz below: five multiple choice questions on terms, sums, geometric means and the AM-GM relation. Attempt them after the notes above, then move to the previous year page.