Chapter 9 – Sequence And Series

Class 11 Mathematics · 29 questions · 0 with answers

Solved examples

example-1Long answer

The first term of an A.P. is a, the second term is b and the last term is c. (b + c − 2a) (c + a) Show that the sum of the A.P. is . 2 (b − a) Solution Let d be the common diffrence and n be the number of terms of the A.P. Since the first term is a and the second term is b Therefore, d= b–a Also, the last term is c, so c = a + (n – 1) (b – a) (since d = b – a) c−a ⇒ n – 1 = b−a c−a b − a + c − a b + c − 2a ⇒ n= 1+ = = b−a b−a b−a n (b + c − 2a) Therefore, (a + l ) = Sn = (a + c) 2 2 (b − a)

example-2Short answer

The pth term of an A.P. is a and qth term is b. Prove that the sum of its (p + q) terms is SEQUENCE AND SERIES 151 p+q a−b a+b+

example-3Long answer

If there are (2n + 1) terms in an A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is (n + 1) : n Solution Let a be the first term and d the common difference of the A.P. Also let S1 be the sum of odd terms of A.P. having (2n + 1) terms. Then S1 = a1 + a3 + a5 + ... + a2n + 1 n +1 S1 = (a1 + a2 n + 1 ) n +1 S1 = [ a + a + (2n + 1 − 1)d ] = (n + 1) (a + nd) Similarly, if S2 denotes the sum of even terms, then S2 = [2a + 2nd] = n (a + nd) S1 (n + 1) (a + nd ) n + 1 Hence = = S2 n (a + nd ) n

example-4Long answer

At the end of each year the value of a certain machine has depreciated by 20% of its value at the beginning of that year. If its initial value was Rs 1250, find the value at the end of 5 years. Solution After each year the value of the machine is 80% of its value the previous year so at the end of 5 years the machine will depreciate as many times as 5. Hence, we have to find the 6th term of the G.P. whose first term a1 is 1250 and common ratio r is .8. Hence, value at the end 5 years = t6 = a1 r5 = 1250 (.8)5 = 409.6

example-5Long answer

Find the sum of first 24 terms of the A.P. a1, a2, a3, ... if it is known that a1 + a5 + a10 + a15 + a20 + a24 = 225. Solution We know that in an A.P., the sum of the terms equidistant from the beginning and end is always the same and is equal to the sum of first and last term. Therefore d= b–a i.e., a1 + a24 = a5 + a20 = a10 + a15 It is given that (a1 + a24) + (a5 + a20) + (a10 + a15) = 225 ⇒ (a1 + a24) + (a1 + a24) + (a1 + a24) =225 ⇒ 3 (a1 + a24) = 225 ⇒ a1 + a24 = 75 We know that Sn = [a + l ] , where a is the first term and l is the last term of an A.P. Thus, S24 = [a1 + a24] = 12 × 75 = 900

example-6Long answer

The product of three numbers in A.P. is 224, and the largest number is 7 times the smallest. Find the numbers. Solution Let the three numbers in A.P. be a – d, a, a + d (d > 0) SEQUENCE AND SERIES 153 Now (a – d) a (a + d) = 224 ⇒ a (a2 – d2) = 224 ... (1) Now, since the largest number is 7 times the smallest, i.e., a + d = 7 (a – d) 3a Therefore, d= Substituting this value of d in (1), we get 9a 2 a a2 − = 224 a= 8 3a 3 and d= = ×8 = 6 4 4 Hence, the three numbers are 2, 8, 14.

example-7Long answer

Show that (x2 + xy + y2), (z2 + xz + x2) and (y2 + yz + z2) are consecutive terms of an A.P., if x, y and z are in A.P. Solution The terms (x2 + xy + y2), (z2 + xz + x2) and (y2 + yz + z2) will be in A.P. if (z2 + xz + x2) – (x2 + xy + y2) = (y2 + yz + z2) – (z2 + xz + x2) i.e., z2 + xz – xy – y2 = y2 + yz – xz – x2 i.e., x2 + z2 + 2xz – y2 = y2 + yz + xy i.e., (x + z)2 – y2 = y (x + y + z) i.e., x+z–y= y i.e., x + z = 2y which is true, since x, y, z are in A.P. Hence x2 + xy + y2, z2 + xz + x2, y2 + yz + z2 are in A.P.

example-8Short answer

If a, b, c, d are in G.P., prove that a2 – b2, b2 – c2, c2 – d2 are also in G.P. Solution Let r be the common ratio of the given G.P. Then = = =r ⇒ b = ar, c = br = ar2, d = cr = ar3 Now, a2 – b2 = a2 – a2r2 = a2 (1 – r2) b2 – c2 = a2r2 – a2r4 = a2r2 (1 – r2) and c2 – d2 = a2r4 – a2r6 = a2r4 (1 – r2) b2 − c2 c2 − d 2 Therefore, = = r2 a 2 − b2 b2 − c2 Hence, a2 – b2, b2 – c2, c2 – d2 are in G.P.

example-9Long answer

If the sum of m terms of an A.P. is equal to the sum of either the next n terms or the next p terms, then prove that 1 1 1 1 (m + n) − = (m + p) − m p m n Solution Let the A.P. be a, a + d, a + 2d, ... . We are given a1 + a2 + ... + am = am+1 + am+2 + ... + am+n ... (1) Adding a1 + a2 + ... + am on both sides of (1), we get 2 [a1 + a2 + ... + am] = a1 + a2 + ... + am + am+1 + ... + am+n 2 Sm = Sm+n m m+n Therefore, 2 {2a + (m −1)d } = {2a + (m + n −1) d } 2 2 Putting 2a + (m – 1) d = x in the above equation, we get m+n mx = (x + nd) (2m – m – n) x = (m + n) nd ⇒ (m – n) x = (m + n) nd ... (2) Similarly, ifa1 + a2 + ... + am = am + 1 + am + 2 + ... + am + p Adding a1 + a2 + ... + am on both sides we get, 2 (a1 + a2 + ... + am) = a1 + a2 + ... + am + 1 + ... + am + p or, 2 Sm = Sm + p m   m+ p ⇒ 2  {2a + (m −1) d } = {2a + (m + p – 1)d} which gives 2  i.e., (m – p) x = (m + p)pd ... (3) Dividing (2) by (3), we get SEQUENCE AND SERIES 155 (m − n) x (m + n) nd = (m − p ) x (m + p ) pd ⇒ (m – n) (m + p) p = (m – p) (m + n) n Dividing both sides by mnp, we get 1 1 1 1 (m + p) − = (m + n) − n m p m 1 1 1 1 = (m + n) − = (m + p) − m p m n

example-10Long answer

If a1, a2, ..., an are in A.P. with common difference d (where d ≠ 0); then the sum of the series sin d (cosec a1 cosec a2 + cosec a2 cosec a3 + ...+ cosec an–1 cosec an) is equal to cot a1 – cot an Solution We have sin d (cosec a1 cosec a2 + cosec a2 cosec a3 + ...+ cosec an–1 cosec an) 1 1 1 = sin d sin a sin a + sin a sin a + ... + sin a sin a 1 2 2 3 n −1 n sin (a2 − a1 ) sin (a3 − a2 ) sin (an − an −1 ) = + + ... + sin a1 sin a2 sin a2 sin a3 sin an −1 sin an sin a2 cos a1 − cos a2 sin a1 ) sin a3 cos a2 − cos a3 sin a2 ) sin an cos an −1 − cos an sin an −1 ) = + + ... + sin a1 sin a2 sin a2 sin a3 sin an −1 sin an = (cot a1 – cot a2) + (cot a2 – cot a3) + ... + (cot an–1 – cot an) = cot a1 – cot an

example-11Multiple choice

(i)If a, b, c, d are four distinct positive quantities in A.P., then show that bc > ad
(ii)If a, b, c, d are four distinct positive quantities in G.P., then show that a+d>b+c Solution (i) Since a, b, c, d are in A.P., then A.M. > G.M., for the first three terms. a +c Therefore, b > ac Here =b Squaring, we get b2 > ac ... (1) Similarly, for the last three terms AM > GM b +d c> bd Here =c c2 > bd ... (2) Multiplying (1) and (2), we get b2c2 > (ac) (bd) ⇒ bc > ad (ii) Since a, b, c, d are in G.P. again A.M. > G.M. for the first three terms a +c >b (since ac = b ) ⇒ a + c > 2b ... (3) Similarly, for the last three terms b +d >c (since bd = c ) ⇒ b + d > 2c ... (4) Adding (3) and (4), we get (a + c) + (b + d) > 2b + 2c a+d>b+c Eample 12 If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. Then prove that xb – c . yc – a . z a – b = 1 Solution We have a, b, c as three consecutive terms of A.P. Then b–a=c–b=d (say) c – a = 2d a–b= –d SEQUENCE AND SERIES 157 Now x b–c . y c–a . z a–b = x – d . y 2d . z – d = x – d ( xz )2 d . z − d (since y = ( xz ) ) as x, y, z are G.P.) = x – d . xd . z d . z – d = x– d + d . zd – d = x° z° = 1

example-13Long answer

Find the natural number a for which f (a + k ) = 16(2n – 1), where k =1 the function f satisfies f (x + y) = f (x) . f (y) for all natural numbers x, y and further f (1) = 2. Solution Given that f (x + y) = f (x) . f (y) and f (1) = 2 Therefore, f (2) = f (1 + 1) = f (1) . f (1) = 22 f (3) = f (1 + 2) = f (1) . f (2) = 23 f (4) = f (1 + 3) = f (1) . f (3) = 24 and so on. Continuing the process, we obtain f (k) = 2k and f (a) = 2a n n Hence f (a + k ) = f (a). f (k ) k =1 k =1 = f (a) f (k ) k =1 = 2a (21 + 22 + 23 + ... + 2n) 2. 2n − 1 ( ) =2 a +1 = 2 (2n −1) ... (1) 2 −1 But, we are given f (a + k ) = 16 (2n – 1) k =1 2a + 1 (2n – 1) = 16 (2n – 1) ⇒ 2a+1 = 24 ⇒ a + 1 = 4 ⇒ a= 3

example-14Multiple choice

A sequence may be defined as a

(A)relation, whose range ⊆ N (natural numbers)
(B)function whose range ⊆ N
(C)function whose domain ⊆ N
(D)progression having real values Solution (C) is the correct answer. A sequence is a function f : N → X having domain ⊆ N

example-15Multiple choice

If x, y, z are positive integers then value of expression (x + y) (y + z) (z + x) is

(A)= 8xyz
(B)> 8xyz
(C)< 8xyz
(D)= 4xyz Solution (B) is the correct answer, since x+ y y+z z+x A.M. > G.M., > xy , > yz and > zx 2 2 2 Multiplying the three inequalities, we get x+ y y+ z y+ z . . > ( xy )( yz )( zx) 2 2 2 or, (x + y) (y + z) (z + x) > 8 xyz

example-16Multiple choice

In a G.P. of positive terms, if any term is equal to the sum of the next two terms. Then the common ratio of the G.P. is

(A)sin 18°
(B)2 cos18°
(C)cos 18°
(D)2 sin 18° Solution (D) is the correct answer, since tn = tn+1 + tn+2 ⇒ arn–1 = arn + arn+1 ⇒ 1 = r + r2 −1 ± 5 r= , since r > 0 5 −1 Therefore, r= 2 = 2 sin 18° SEQUENCE AND SERIES 159

example-17Multiple choice

In an A.P. the pth term is q and the (p + q)th term is 0. Then the qth term is

(A)– p (B p
(B)is the correct answer Let a, d be the first term and common difference respectively. Therefore, Tp = a + (p – 1) d = q and ... (1) Tp+ q = a + (p + q – 1) d = 0 ... (2) Subtracting (1), from (2) we get qd = – q Substituting in (1) we get a = q – (p – 1) (–1) = q + p – 1 Now Tq = a + (q – 1) d = q + p – 1 + (q – 1) (–1) = q+p–1–q+1=p
(C)p + q
(D)p – q Solution

example-18Multiple choice

Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then P2 R3 : S3 is equal to

(A)1 : 1
(B)(common ratio)n : 1
(C)(first term)2 : (common ratio)2
(D)none of these Solution (A) is the correct answer Let us take a G.P. with three terms , a, ar . Then a a (r 2 + r + 1) S= + a + ar = r r r 1 1 1 r2 + r + 1 P = a3, R = + + = a a ar a r

example-19Multiple choice

The 10th common term between the series 3 + 7 + 11 + ... and 1 + 6 + 11 + ... is

(A)191
(B)193
(C)211
(D)None of these Solution (A) is the correct answer. The first common term is 11. Now the next common term is obtained by adding L.C.M. of the common difference 4 and 5, i.e., 20. Therefore, 10th common term = T10 of the AP whose a = 11 and d = 20 T10 = a + 9 d = 11 + 9 (20) = 191

example-20Multiple choice

In a G.P. of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the G.P. is −4 1

(A)
(B)
(C)4
(D)none the these 5 5 Solution (C) is the correct answer 2 n a (r 2 n −1) 5a ( (r ) −1) Let us consider a G.P. a, ar, ar , ... with 2n terms. We have = r −1 r 2 −1 (Since common ratio of odd terms will be r2 and number of terms will be n) a (r 2 n −1) a (r 2 n −1) ⇒ = 5 r −1 (r 2 −1) ⇒ a (r + 1) = 5a, i.e., r = 4

example-21Multiple choice

The minimum value of the expression 3x + 31 – x, x ∈ R, is

(A)0
(B)
(C)3
(D)2 3 Solution (D) is the correct answer. We know A.M. ≥ G.M. for positive numbers. 3x + 31− x Therefore, ≥ 3x ⋅ 31− x 3x + 31− x 3 ⇒ ≥ 3x ⋅ x 2 3 ⇒ 3x + 31–x ≥ 2 3 SEQUENCE AND SERIES 161

Questions

Q2Short answer

p−q . Solution Let A be the first term and D be the common difference of the A.P. It is given that tp = a ⇒ A + (p – 1) D = a ... (1) tq = b ⇒ A + (q – 1) D = b ... (2) Subtracting (2) from (1), we get (p – 1 – q + 1) D = a – b a−b ⇒ D= p−q ... (3) Adding (1) and (2), we get 2A + (p + q – 2) D = a + b ⇒ 2A + (p + q – 1) D = a + b + D a−b ⇒ 2A + (p + q – 1) D = a + b + p − q .. (4) p+q Now Sp + q = [2A + (p + q – 1) D] p+q a−b = a+b+ 2 p−q [(using ... (3) and (4)]

Q6Multiple choice

1 r2 + r + 1 a ⋅ 3 P2 R 3 a r =1 = 3 S3 r2 + r + 1 a3 Therefore, the ratio is 1 : 1

Q1Short answer

The first term of an A.P.is a, and the sum of the first p terms is zero, show that −a ( p + q) q the sum of its next q terms is . [Hint: Required sum = Sp + q – Sp] p −1

Q2Short answer

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?

Q3Multiple choice

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?

Q4Short answer

If the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)th qp p−q term is .

Q5Short answer

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?

Q6Short answer

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.

Q7Short answer

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.

Q8Short answer

In a potato race 20 potatoes are placed in a line at intervals of 4 metres with the first potato 24 metres 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?

Q9Short answer

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?

Q10Short answer

If a1, a2, a3, ..., an are in A.P., where ai > 0 for all i, show that 1 1 1 n −1 + + ... + = a1 + a2 a2 + a3 an −1 + an a1 + an

Q11Multiple choice

Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + ... to

(i)n terms
(ii)10 terms

Q12Short answer

Find the rth term of an A.P. sum of whose first n terms is 2n + 3n2. [Hint: an = Sn – Sn–1]

Q13Long answer

If A is the arithmetic mean and G1, G2 be two geometric means between any two numbers, then prove that G12 G 22 2A = + G 2 G1

Q14Long answer

If θ1, θ2, θ3, ..., θn are in A.P., whose common difference is d, show that tan θn − tan θ1 secθ1 secθ2 + secθ2 secθ3 + ... + secθn–1 secθn = . sin d

Q15Long answer

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).

Q16Long answer

If pth, qth, and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab–c . bc – a . ca – b = 1

Q26Multiple choice

(M.C.Q.). 17. If the sum of n terms of an A.P. is given by Sn = 3n + 2n2, then the common difference of the A.P. is

(A)3
(B)2
(C)6
(D)4 SEQUENCE AND SERIES 163 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 19. If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then the 22nd term of the A.P. is (A) 0 (B) 22 (C) 220 (D) 198 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 1 1 (A) 3 (B) (C) 2 (D) 3 2 21. If in an A.P., Sn = q n2 and Sm = qm2, where Sr denotes the sum of r terms of the A.P., then Sq equals q3 (A) (B) mnq (C) q 3 (D) (m + n) q2 22. Let Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn then S3n : Sn is equal to (A) 4 (B) 6 (C) 8 (D) 10 x 1–x 23. The minimum value of 4 + 4 , x ∈ R, is (A) 2 (B) 4 (C) 1 (D) 0 24. Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then equals r =1 sr n(n + 1)(n + 2) n(n + 1) (A) (B) 6 2 n 2 + 3n + 2 (C) (D) None of these 25. If tn 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 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 252cm2. The length of the longest edge is (A) 12 cm (B) 6 cm (C) 18 cm (D) 3 cm Fill in the blanks in the Exercises 27 to 29. a −b

Q27Multiple choice

For a, b, c to be in G.P. the value of b − c is equal to .............. .

Q28Multiple choice

The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .

Q29True / False

The third term of a G.P. is 4, the product of the first five terms is ................ . State whether statement in Exercises 30 to 34 are True or False.

Q30Multiple choice

Two sequences cannot be in both A.P. and G.P. together.

Q31Multiple choice

Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.

Q32Multiple choice

Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it.

Q33Multiple choice

The sum or difference of two G.P.s, is again a G.P.

Q34Match the following

If the sum of n terms of a sequence is quadratic expression then it always represents an A.P. Match the questions given under Column I with their appropriate answers given under the Column II.

Q35Multiple choice

Column I Column II 1 1 (a) 4, 1, ,

(i)A.P. 4 16 (b) 2, 3, 5, 7
(ii)sequence (c) 13, 8, 3, –2, –7
(iii)G.P.

Q36Multiple choice

Column I Column II n( n + 1) (a) 12 + 22 + 32 + ...+n2

(i)(b) 13 + 23 + 33 + ...+n3
(ii)n (n + 1) n (n +1)(2n +1) (c) 2 + 4 + 6 + ... + 2n
(iii)n (n +1) (d) 1 + 2 + 3 +...+ n
(iv)