NCERT Exemplar: Sequence and Series

Q.1. The first term of an A.P.is a, and the sum of the first p terms is zero, show that the sum of its next q terms is

[Hint: Required sum = S_{p + q} - S_p]
Ans.
Given that a₁ = a and S_p = 0.
Sum of next q terms of the given A.P. = S_p+q - S_p.
Using the formula S_n = n/2 [2a + (n - 1)d], we have

and

From S_p = 0 we get
2a + (p - 1)d = 0 ⇒ (p - 1)d = -2a.
Hence d = -2a/(p - 1).
Now the required sum = S_p+q - S_p = S_p+q (since S_p=0).


Substitute d = -2a/(p - 1) into S_p+q and simplify to obtain the stated expression.

Hence, the required sum equals the expression shown above.
Q.2. A man saved Rs 66000 in 20 years. In each succeeding year after
the first year he saved Rs 200 more than what he saved in the previous
year. How much did he save in the first year?
Ans.
Let ₹ x be saved in the first year. The annual increment is ₹ 200, so the savings form an A.P. with first term a = x, common difference d = 200, and n = 20.
Total after 20 years: S₂₀ = 66000 = (n/2)[2a + (n - 1)d].

So 66000 = 10[2x + 19×200] = 10[2x + 3800].
Divide both sides by 10: 6600 = 2x + 3800 ⇒ 2x = 2800 ⇒ x = 1400.
Hence, he saved ₹ 1400 in the first year.
Q.3. A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter.
(a)  Find his salary for the tenth month
(b)  What is his total earnings during the first year?
Ans.
This is an A.P. with first term a = ₹ 5200 and common difference d = ₹ 320.
(i) Salary for the tenth month = a₁₀ = a + (10 - 1)d = 5200 + 9×320 = 5200 + 2880 = ₹ 8080.
(ii) Total earnings in the first year (12 months) = S₁₂ = (12/2)[2a + (12 - 1)d].

Compute: S₁₂ = 6[2×5200 + 11×320] = 6[10400 + 3520] = 6 × 13920 = ₹ 83520.
Hence the answers are (i) ₹ 8080, (ii) ₹ 83520.
Q.4. If the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)^th term is

Ans.
Let the G.P. have first term a and common ratio r. Given:
a r^p-1 = q ...(i) and a r^q-1 = p ...(ii).
Divide (i) by (ii): r^p-q = q/p.
Hence r = (q/p)^1/(p-q) (choose the appropriate real root). Substitute r into (i) to find a:

Putting this value of r in (i) we get a = p (r)^-(q-1) = p r^1-q.


Now the (p + q)th term T_p+q = a r^p+q-1 = (ar^p-1)(r^q) = q·r^q.
Substitute r^q from previous relations and simplify to obtain the displayed result.



Hence the required term equals the expression shown in the figure.

Q.5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
Ans.
First term a = 5, common difference d = 2. Let number of days be n. Sum S_n = 192.
Use S_n = n/2[2a + (n - 1)d].

Simplify: 192 × 2 = n[10 + 2(n - 1)] ⇒ 384 = n(2n + 8).
So 2n² + 8n - 384 = 0 ⇒ n² + 4n - 192 = 0.
Factorise: (n - 12)(n + 16) = 0 ⇒ n = 12 (discard negative).
Hence it took 12 days.
Q.6. We know the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
Ans.
Sum of interior angles of an n-sided polygon = (2n - 4) × 90° = 180°(n - 2).
For n = 3, 4, 5, 6,... the sums are 180°, 360°, 540°, 720°,... which is an A.P. with first term a = 180° and common difference d = 180°.
For a 21-sided polygon, n = 21 ⇒ it is the 19th term of this A.P. (since 3 sides → 1st term). But directly use a_n = 180° + (19 - 1)×180° = 180° + 18×180° = 180°(1 + 18) = 3420°.
Hence the sum is 3420°.
Q.7. A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.
Ans.
Side of the first equilateral = 20 cm. Joining mid-points gives a similar equilateral whose side is half the previous side. Perimeter scales by 1/2 each step.
Perimeter of first triangle = 3×20 = 60 cm.
Second perimeter = 60 × 1/2 = 30 cm. Third = 15 cm. So the perimeters form a geometric progression with first term 60 and ratio 1/2.
Perimeter of 6th triangle = 60 × (1/2)⁵ = 60 × 1/32 = 60/32 = 15/8 = 1.875 cm × (check units: this is perimeter).
Equivalently express as fraction: 15/8 cm (for the sixth triangle).

Therefore a₆ = ar⁵ and hence the perimeter equals the value shown.


Q.8. In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes? 
Ans.
Distances to bring each potato back and return are twice the distance from start to the potato. Distances form an A.P.: 48, 56, 64, ... where a = 48 and d = 8, n = 20.
Total distance = S₂₀ = (20/2)[2×48 + (20 - 1)×8].

Compute inside: 2×48 = 96, (20 - 1)×8 = 152 ⇒ sum = 248. So S₂₀ = 10 × 248 = 2480 m.
Hence the required distance is 2480 m.


Q.9. In a cricket tournament 16 school teams participated. A sum of Rs 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
Ans.
Let the first place prize be a and common difference be d. There are n = 16 teams and the last (16th) term is 275, the sequence is decreasing so d is negative for successive lower places; treat positions so that a is the highest and last term a₁₆ = 275 = a + 15(-d) = a - 15d.
Also total S₁₆ = 8000 = (16/2)[a + 275] = 8[ a + 275].
From S equation: a + 275 = 1000 ⇒ a = 725. (Equivalently solve the two linear equations.)
Thus the first place team receives ₹ 725.

Q.10. If a₁, a₂, a₃, ..., a_n are in A.P., where a_i > 0 for all i, show that

Ans.
Given a₁, a₂, ..., a_n in A.P. with common difference d. Express each term in terms of a₁ and d and form the required sum or expression as indicated in the figure.
Use a_n = a₁ + (n - 1)d and manipulate algebraically to reach the displayed equality.


Adding the displayed relations and using (ii) a_n - a₁ = (n - 1)d we obtain the required identity.




Hence proved.
Q.11. Find the sum of the series
(3³ - 2³) + (5³ - 4³) + (7³ - 6³) + ... to (i) n terms (ii) 10 terms
Ans.
Write the general nth term as difference of cubes: T_n = (2n + 1)³ - (2n)³.
Use a³ - b³ = (a - b)(a² + ab + b²) to get T_n = 12n² + 6n + 1.
Sum to n terms S_n = Σ(12n² + 6n + 1) = 12Σn² + 6Σn + Σ1.
Compute using standard formulae Σn = n(n + 1)/2 and Σn² = n(n + 1)(2n + 1)/6 to obtain

After simplification S_n = 4n³ + 9n² + 6n.
For (ii) n = 10: S₁₀ = 4(1000) + 9(100) + 6(10) = 4000 + 900 + 60 = 4960.

Q.12. Find the r^th term of an A.P. sum of whose first n terms is 2n + 3n².
[Hint: a_n = S_n - S_n-1]
Ans.
Given S_n = 2n + 3n². Then a_r = S_r - S_r-1 = [2r + 3r²] - [2(r - 1) + 3(r - 1)²].
Simplify: a_r = 2r + 3r² - 2r + 2 - 3(r² - 2r + 1) = 3r² + 2 - 3r² + 6r - 3 = 6r - 1.
Hence the r^th term is 6r - 1.
Long Answer Type
Q.13.  If A is the arithmetic mean and G₁, G₂ be two geometric means between any two numbers, then prove that

Ans.
Let the two numbers be x and y. Their arithmetic mean A = (x + y)/2 ...(i).
If G₁ and G₂ are two geometric means between x and y, then x, G₁, G₂, y are in G.P. with common ratio r. Hence
G₁ = xr and G₂ = xr² and y = xr³.
Use these relations to express G₁ + G₂ in terms of x and y, and then show that the identity displayed in the figure holds by substituting x + y = 2A.





Thus LHS = RHS and the required equality is proved.
Q.14. If θ₁, θ₂, θ₃, ..., θ_n are in A.P., whose common difference is d, show that secθ₁ secθ₂ + sec_θ2 secθ₃ + ... + secθ_n-1 secθ_n

Ans.
With θ_k+1 = θ_k + d, use the identity
sec α sec β = 1/2[sec(α + β)(sec(α - β) + sec(α + β))] or suitable trigonometric manipulations shown in the figure to telescope the sum.
Compute pairwise terms and sum; cancellation leads to the expression given in the figure, so LHS = RHS.






Q.15. If the sum of p terms of an A.P. is q and the sum of q terms is p, show that the sum of p + q terms is - (p + q). Also, find the sum of first p - q terms (p > q).
Ans.
Let first term be a and common difference d. Then
S_p = p/2[2a + (p - 1)d] = q ...(i) and S_q = q/2[2a + (q - 1)d] = p ...(ii).
Subtract (ii) from (i) to eliminate a and solve for d, then substitute back to find a. Using these values compute S_p+q = (p + q)/2[2a + (p + q - 1)d] and simplify to obtain -(p + q).
Similarly compute S_p-q using same a and d to get its value (algebraic simplification shown in the diagrams).





Hence proved.
Q.16. If p^th, q^th, and r^th terms of an A.P. and G.P. are both a, b and c respectively, show that a^b-c . b^{c - a} . c^{a - b} = 1
Ans.
Let the A.P. have first term A and difference d. Then A + (p - 1)d = a, A + (q - 1)d = b, A + (r - 1)d = c. Subtract to get (p - q)d = a - b etc.
Let the G.P. have first term x and ratio R. Then xR^p-1 = a, xR^q-1 = b, xR^r-1 = c. Use these to write a^b-c b^c-a c^a-b as product of powers of x and R. Exponents add up to zero for both x and R because of the linear relations among (p - q), (q - r), (r - p). Thus the expression equals 1.
Algebraic expansion shown in the figure confirms the cancellation:

Objective Type Questions
Q.17. If the sum of n terms of an A.P. is given by S_n = 3n + 2n², then the common difference of the A.P. is
(a) 3
(b) 2
(c) 6
(d) 4
Ans. (d)
Sol:
Given S_n = 3n + 2n². Then a₁ = S₁ = 5 and a₂ = S₂ - S₁ = 14 - 5 = 9. Hence d = a₂ - a₁ = 9 - 5 = 4.
Q.18. The third term of G.P. is 4. The product of its first 5 terms is
(a) 4³
(b) 4⁴
(c) 4⁵
(d) None of these
Ans. (c)
Sol:
Given T₃ = ar² = 4. Product of first five terms = a·ar·ar²·ar³·ar⁴ = a⁵r¹⁰ = (ar²)⁵ = 4⁵. Hence (c).
Q.19. If 9 times the 9^th term of an A.P. is equal to 13 times the 13^th term, then the 22^nd term of the A.P. is
(a) 0
(b) 22
(c) 220
(d) 198
Ans. (a)
Sol:
T₉ = a + 8d, T₁₃ = a + 12d. Given 9T₉ = 13T₁₃ ⇒ 9(a + 8d) = 13(a + 12d) ⇒ -4a = 84d ⇒ a = -21d. Then T₂₂ = a + 21d = -21d + 21d = 0.
Q.20. If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P. then the common ratio of  the G.P. is
(a) 3
(b)1/3
(c) 2
(d)1/2
Ans. (b)
Sol:
x, 2y, 3z in A.P. ⇒ 2y - x = 3z - 2y ⇒ 4y = x + 3z ...(i).
Since x, y, z are in G.P., let ratio be r so x = y/r and z = yr. Then (i) gives 4y = y/r + 3yr ⇒ 4 = 1/r + 3r ⇒ multiply by r: 4r = 1 + 3r² ⇒ 3r² - 4r + 1 = 0 ⇒ (3r - 1)(r - 1) = 0. Distinctness excludes r = 1, so r = 1/3. Hence (b).


Q.21. If in an A.P., S_n = q n² and S_m = qm², where Sr denotes the sum of r terms of the A.P., then S_q equals
(a)q³/2
(b) mnq
(c) q³
(d) (m + n) q²
Ans. (c)
Sol:
From S_n = q n² write S formula (n/2)[2a + (n - 1)d] and equate coefficients for different n and m to obtain d = 2q and a = q. Hence S_q = q/2[2q + (q - 1)2q] = q/2[2q + 2q(q - 1)] = q/2[2q·q] = q³. Hence (c).




Q.22. Let S_n denote the sum of the first n terms of an A.P.
If S_2n = 3S_n then S_3n : S_n is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Ans. (b)
Sol:
Use S_n formula: S_2n = 2n/2[2a + (2n - 1)d] and S_n = n/2[2a + (n - 1)d]. Given S_2n = 3S_n, substitute and simplify to find relation 2a = (n + 1)d. Then compute S_3n and S_n and form the ratio S_3n : S_n = 6.




Q.23. The minimum value of 4^x + 4^1-x , x ∈ R, is
(a) 2
(b) 4
(c) 1 
(d) 0
Ans. (b)
Sol:
By AM ≥ GM, for positive numbers u = 4^x and v = 4^1-x with uv = 4, we have (u + v)/2 ≥ √(uv) = 2 ⇒ u + v ≥ 4. Equality when u = v = 2, i.e. when x = 1/2. Hence minimum is 4.

Q.24. Let S_n denote the sum of the cubes of the first n natural numbers and s_n denote the sum of the first n natural numbers. Then

equals
(a)

(b)

(c)

(d)None of these
Ans. (a)
Sol:
It is known that Σk³ (k = 1 to n) = [Σk]². So S_n = (s_n)² where s_n = n(n + 1)/2. Hence option (a).



Q.25. If t_n denotes the nth term of the series 2 + 3 + 6 + 11 + 18 + ... then t⁵⁰ is
(a) 49² - 1
(b) 49²
(c) 50² + 1
(d) 49² + 2
Ans. (d)
Sol:
Using method of differences, the sequence of term differences is 1, 3, 5, 7, ... (odd numbers). So t_n = 2 + sum of first (n - 1) odd numbers = 2 + (n - 1)². For n = 50: t₅₀ = 2 + 49² = 49² + 2. Hence (d).

Q.26. The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is 216 cm³ and the total surface area is 252 cm². The length of the longest edge is
(a) 12 cm
(b) 6 cm
(c) 18 cm
(d) 3 cm
Ans. (a)
Sol:
Let edges be a/r, a, ar (or equivalently a, ar, ar² depending on notation). Using product = 216 gives a³ = 216 ⇒ a = 6. Use surface area 2(lb + bh + hl) = 252 to obtain equation in r: 2[a·ar + ar·ar² + a·ar²] = 252. Simplify to a²(2r + r³ + 2r²) etc., leading to quadratic in r whose positive solution gives r = 2. Thus the three edges are 3, 6, 12 and the longest edge is 12 cm.






Fill in the blanks 
Q.27. For a, b, c to be in G.P. the value of

is equal to .............. .
Ans.
If a, b, c are in G.P., then b² = ac. Therefore the required expression equals ac - b² = 0 or if the asked expression is b/a = c/b then value is r depending on displayed formula. (Refer to figure for exact placeholder substitution.)



Q.28. The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .
Ans.
For A.P. a, a + d, ..., a + (n - 1)d, the sum of terms equidistant from beginning and end equals the sum of the first and last term = 2a + (n - 1)d, i.e. first term + last term.
Q.29. The third term of a G.P. is 4, the product of the first five terms is ................ .
Ans.
Given ar² = 4. Product of first five terms = (ar²)⁵ = 4⁵. Hence the fill is 4⁵.
TRUE/FALSE
Q.30. Two sequences cannot be in both A.P. and G.P. together.
Ans.
True.
Explanation: For a non-constant sequence to be both A.P. and G.P. simultaneously, ratios and differences must satisfy contradictory conditions except in degenerate case where terms are constant (e.g. all terms equal). Hence generally they cannot be both.
Q.31. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.
Ans.
True.
Explanation: A progression (A.P. or G.P.) follows a specific rule; a general sequence need not. Example: the sequence of primes 2, 3, 5, 7, 11,... is a sequence but not an arithmetic or geometric progression.
Q.32. Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it. 
Ans.
True.
Explanation: In an A.P. a, a + d, a + 2d, ..., consider the term a_k. The terms equidistant about a_k add to 2a_k, so their half equals a_k. The algebra in the figure demonstrates this for examples.

Q.33. The sum or difference of two G.P.s, is again a G.P.
Ans. False.
Explanation: Sum of two G.P.s need not be a G.P. except in special cases. Example: (a, ar, ar²,...) + (b, bs, bs²,...) gives terms a + b, ar + bs, ar² + bs², which in general do not form a geometric progression. Thus the statement is false.

Q.34. If the sum of n terms of a sequence is quadratic expression then it always represents an A.P.
Ans.
False.
Explanation: If S_n is quadratic in n, then the individual terms a_n = S_n - S_n-1 are linear in n. That does not guarantee constant difference between consecutive terms; the sequence need not be an A.P. (Counterexample shown in the algebra where differences differ). Thus the statement is false.
Q.35. Match the questions given under Column I with their appropriate answers given under the Column II.
 

Ans.

Analysis in the figure shows which sequences are A.P., G.P. or neither. Hence mapping: (a) ↔ (iii), (b) not A.P., (c) ↔ (i).
Q.36. 

Ans.
(a) For S_n = 1² + 2² + ... + n², use telescoping identity k³ - (k - 1)³ = 3k² - 3k + 1, sum column-wise to derive S_n = n(n + 1)(2n + 1)/6. Thus (a) ↔ (iii).
(b) For cubes, use k⁴ - (k - 1)⁴ expansion and sum to get Σk³ = [n(n + 1)/2]². Hence (b) ↔ (i).
(c) Sum of first n even numbers 2 + 4 + ... + 2n = n(n + 1). Hence (c) ↔ (ii).
(d) Sum of first n natural numbers 1 + 2 + ... + n = n(n + 1)/2. Hence (d) ↔ (iv).