PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1zckRnE4-zPpJLGHHpF8aEvi5o_4vFb-n/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true Sequences and Series Arithmetic and Geometric progressions. Insertion of Arithmetic, Geometric means between two given numbers. Relation between A.M. and G.M. Sum upto n terms of Special Series : Σn, Σn2, Σn3 . Arithmetico-Geometric progression. SEQUENCE, PROGRESSION AND SERIES A sucession of numbers t1, t2, ... tn formed according to the some definite rule is called sequence. “A sequence is a function of natural numbers with codomain as the set of Real numbers or complex numbers” Sequence is called finite or infinite depending upon its having number of terms as finite or infinite respectively. For example: 2, 3, 5, 7, 11, .... is a sequence of prime numbers. It is an infinite sequence. A progression is a sequence having its terms in a definite pattern e.g.: 1, 4, 9, 16, is a progression as each successive term is obtained by squaring the next natural number. However a sequence may not always have an explicit formula of nth term. Series is constructed by adding or subtracting the terms of a sequence e.g., 2 + 4 + 6 + 8 + ..... is a series. The term at nth place is denoted by T and is called general term of a sequence or progression or series. ARITHMETIC PROGRESSION (A.P.) It is sequence in which the difference between any term and its just preceding term remains constant throughout. This constant is called the “common difference” of the A.P. and is denoted by ‘d ’ generally. A.P. is of the form a,(a + d ), (a + 2d ).... where ‘a’ denotes the first term or initial term C H A P T E R CHAPTER INCLUDES : Sequence, Progression and series Arithmetic progression Geometric progression Means Special series Arithmetico- Geometric series Method of difference Solved examples Important relations : an – an–1 = d = common difference an = nth term of A.P. = {a + (n – 1) d} = l a′r = rth term of A.P. from the end = (n – r + 1)th term from beginning n = total number of terms i.e., a′r = a(n-r+1) = a + (n – r) d a′n = nth term of A.P. from the end = {l – (n – 1) d} Sn = the sum of first n terms of A.P. = n [2a + (n − 1)d ] = n [a + l ] 2 2 = n [2l − (n − 1)d ] 2 an = Sn – Sn–1 Properties of Arithmetic Progressions If a1, a2, a3, , an are in A.P., then a1 + k, a2 + k ,...., an + k are also in A.P. ka1, ka2 ,...., kan are also in A.P. (c) a1 , a2 , , an ,k ≠ 0 are also in A.P. k k k If a1, a2, a3,...., and b1, b2, b3, are two A.Ps., then a1 + b1, a2 + b2, a3 + b3 , , are also in A.P. a1 – b1, a2 – b2, a3 – b3 , , are also in A.P. If a1, a2, a3, an, are in A.P., then (a) a1 + an = a2 + an–1 = a3 + an–2 = = 2a1 + (n–1)d (b) a = ar −k + ar +k ,0 ≤ k ≤ n − r r 2 If nth term of a sequence is a linear expression in n then the sequence is an A.P. If the sum of first n terms of a sequence is a quadratic expression in n, then the sequence is an A.P. Three numbers a, b, c are in A.P. if and only if b – a = c – b, i.e., if and only if a + c = 2b. Any three numbers in an A.P. can be taken as a – d, a, a + d. Any four numbers in an A.P. can be taken as a – 3d, a – d, a + d, a + 3d. Similarly 5 numbers in A.P. can be taken as a – 2d, a – d, a, a + d, a + 2d. Find six numbers in A.P., such that the sum of the two extremes be 16 and the product of the two middle terms be 63. Solution : Let α – 5β, α – 3β, α – β, α + β, α + 3β, α + 5β be the numbers. By the given conditions, we have : (α – 5β) + (α + 5β) = 16 ; ∴ α = 8 (α – β) (α + β) = 63 ; ∴ β = ± 1 The six numbers are 3, 5, 7, 9, 11, 13. The interior angles of a polygon are in arithmetic progression. The smallest angle is 120° and the common difference is 5°. Find the number of sides of the polygon. Solution : Let the number of sides of the polygon be n. The sum of interior angles of the polygon = (n –2) π = (n – 2) × 180° (1) Also the first term of the A.P. = a = 120° and common difference = d = 5° ∴ Sum of Interior angles = = ⇒ n2 – 25n + 144 = 0 ⇒ (n – 9) (n – 16) = 0 ⇒ n = 9 or 16 n [2.120° + (n − 1)5°] = (n − 2) ×180° 2 n (5n + 235) = (n − 2)180 2 {from (1)} For n = 16. Then T13 = a + (13 – 1) d = 120° + 12 × 5° = 180° but no angle of a polygon is 180°. Hence n = 9 GEOMETRIC PROGRESSION (G.P.) A sequence (finite or infinite) of non-zero numbers in which every term, except the first one, bears a constant ratio with its preceding term, is called a geometric progression. The constant ratio, also called the common ratio of the G.P. is usually denoted by r. For example, in the sequence, 1, 2, 4, 8, .... 2 = 4 = 8 = = 2 , which is a constant. 1 2 4 Thus, the sequence is a G.P. whose first term is 1 and the common ratio is 2. Important Relations For a G.P.; a, ar, ar2, ... arn–1 Properties of Geometrical Progression a1, a2, a3, are in G.P. then a k, a k, a k, ... and a /k, a /k, a /k are also in G.P. (k ≠ 0) If a1, a2, a3, .... are in G.P. Then 1/a1, 1/a2, 1/a3, are also in G.P. If a1, a2, a3, .... and b1, b2, b3, ... are two G.P.s, then a1b1, a2b2, a3b3, ... and a1/b1, a2/b2, a3/b3, are also in G.P. If a , a , a , .... and b , b , b , ... are two G.P.s, then a ± b , a ± b , a ± b , are not in G.P. If a1, a2, a3, .... are in G.P. (ai > 0, ∀i ) , then log a1, log a2, log a3, are in A.P. In this case the converse also holds good. If a1, a2, a3, an are in G.P. , then a a = a a = a a = = a2 rn–1 (b) ar = ar −k ar +k , 0 ≤ k ≤ n − r If a1, a2, a3, a4, , an–1, anare in G.P. then a2 = a3 = a4 = = an a1 a2 a3 an −1 ⇒ a2 = a a , a2 = a a ,... 2 3 1 3 2 4 also a = a r, a = a r2, a = a r3 , ...., a = a rn–1 where r is the common ratio. a a a 2 Three numbers in G.P. can be taken as r , a,ar ; Five numbers in G.P. can be taken as r 2 , r ,a, ar ,ar In general: (2m + 1) numbers in G.P. can be written as (m ∈ N) etc. a , a a m−1 m r m r m−1 ,..., r ,a ar , , ar ,ar a a 3 Four numbers in G.P. can be taken as r 3 , r ,ar,ar ; Six numbers in G.P. can be taken as a , a a 3 5 r 5 r 3 , r ,ar ,ar , ar ; etc. In general: (2m) numbers in G.P. can be written as (m ∈ N) a , a r 2m −1 r 2m−3 ,..., a , a ar ,ar 3, , ar 2m−3,ar 2m−1 r 3 r Find three numbers in G.P. whose sum is 52 and the sum of their products in pairs is 624. Solution : Let three numbers be a, ar, ar2. It is given that a + ar + ar2 = 52 ; ∴ a(1 + r + r2) = 52 ...(1) Also a2r (1 + r + r2) = 624 …(2) a2r (1+ r + r 2 ) = From (1) and (2), we have a2(1+ r + r 2 )2 ⇒ 13r = 3 + 3r + 3r2 ⇒ r = 3, 1/3 624 = 3 52 × 52 13 From (1),a = 4 when r = 3 and a = 36 when r = 1/3 ∴ Numbers are 4, 12, 36. Find the value of 0.325&8& Solution : Let R = 0.325&8& ⇒ R = 0.32585858 ...(1) then 100 R = 32.585858 ...(2) and 10000 R = 3258.5858 ...(3) Subtracting (2) from (3), we get 9900 R = 3226 ∴ Hence R = 3226 9900 R = 1613 4950 RECOGNIZATION OF AP & GP If a, b, c are three successive terms of a sequence If a − b = a = 1 , then a, b, c are in A.P. b − c a If a − b = a , then a, b, c are in G.P. b − c b MEANS Arithmetic Mean If three terms ‘a, b, c’ are in A.P., then middle term ‘b’ is called A.M. between the other two i.e., a and c b = a + c 2 i.e., A.M. of two numbers x1 and x2 is x1 + x2 2 A.M. of n positive numbers = a1 + a2 + a3 + an n Insertion of n Arithmetic means between two numbers Let A1, A2, ... An are n A.M. between a and b then a, A1, A2, ....., An, b form an A.P. b is (n + 2)th term ∴ b = a + (n + 1)d ⇒ d = (b − a) n + 1 Geometric Mean If three terms a, b, c are in G.P., then b is called G.M. of a and c such that G.M. of n numbers = n a1.a2.a3 an Insertion of n G.M. between two numbers (a and b) Here a, G1, G2, , Gn, b will be in G.P. So b = (n + 2)th term of G.P. Hence, b = a . rn+1 ⇒ 1 r = ⎛ b ⎞ n+1 ⎝ a ⎠ k G = ar k = a ⎛ b ⎞ n +1 k a ⎝ ⎠ Relations between A.M. and G.M. For two real positive numbers a and b A = a + b , G = 2 A > G if a ≠ b ...(i) A = G if a = b ...(ii) So combining (i) & (ii), we get A ≥ G, the equality holds when a = b If a1, a2, a3... an are n positive numbers, then Above discussion leads to the result that, a1 + a2 + a3 + .... + an ≥ n a a a a n 1 2 3 n There are n arithmetic means between 1 and 31, such that the 7th mean : (n – 1)th mean ≡ 5 : 9. Find n ? Solution : Let d and Aj denote the common difference and jth Arithmetic mean respectively ; then, d = 31− 1 = n + 1 30 n + 1 A7 = 1+ 7 30 n + 1 = 1+ 210 n + 1 An – 1 = 1+ (n − 1) 30 n + 1 A7 = 5 ⇒ 9 + 1890 = 5 + 150(n − 1) An−1 9 n + 1 n + 1 ⇒ 150n − 150 − 1890 = 4 n + 1 ⇒ 146 n = 2044 ⇒ n = 14. If one A.M., A and two G.M.s’ p and q be inserted between any two given numbers then show that p3 + q3 = 2A pq. Solution : Let the two given numbers be a and b ; then, 2A = a + b ...(1) a, p, q, b are in G.P. ⇒ p2 = aq and q2 = bp ⇒ p3 = apq and q3 = bpq ∴ p3 + q3 = (a + b) pq = 2A pq Special Series Sigma (Σ) notation Σ indicates sum i.e., ∑i = ∑n = 1+ 2 + 3 + + n i =1 n i + 1 1+ 1 2 + 1 3 + 1 n + 1 (i) ∑ i =1 m = + + + .... + i + 2 1+ 2 2 + 2 3 + 2 n + 2 (ii) ∑ i =1 a = a + a + ..... + a m times = am where a is constant (iii) ∑ i =1 ai = a ∑ i = a(1+ 2 + + m) i =1 m m m (iv) ∑ (i 3 − 2i 2 + i ) = ∑ i 3 −2∑ i 2 + ∑ i i =1 i =1 i =1 i =1 3.5 + 6.8 + 9.11 + + upto n terms Solution : nth term of 3, 6, 9, is 3n nth term of 5, 8, 11, is (3n + 2) ∴ T = 3n (3n + 2) = 9n2 + 6n ∴ S = 9Σn2 + 6Σn = 9n(n + 1)(2n + 1) + 6n(n + 1) 6 2 = 3 n(n + 1)[2n + 1+ 2] 2 3n(n + 1)(2n + 3) = 2 ARITHMETICO-GEOMETRIC SERIES (A.G. S.) nth term of A.G.. S. = (nth term of an A.P.) × (nth term of a G.P.) If a, (a + d), (a + 2d) + be an A.P. & b, br, br2 + be a G.P. then ab + (a + d) br + (a + 2d)br2 + is the corresponding A.G.S. Sum of finite A-G series ⇒ S = a 1− r + dr (1− r n−1) (1− r )2 − (a + (n − 1)d )r n (1− r ) For infinite A.G. series ∴ S∞ = a 1− r dr (1− r )2 (| r |< 1) . Find the sum of n terms of the following series 1+ 4 + 7 5 52 10 + upto n terms. 53 Solution : It is an A.G. series with a = 1, d = 3, r = 1 . 5 1 3 ⋅ ⎛ 1 ⎞n 3 ⎜ 5 ⎟ ⎛ 1 ⎞n (3n − 2) ⎜ ⎟ 5 Hence, Sn = + 5 − ⎝ 1− 1 ⎛ 1 ⎞2 ⎛ ⎜1− ⎟ 5 ⎜1− ⎝ ⎠ ⎝ Method of Difference When the difference (or difference of differences) of the successive terms of series are in A.P. or G.P, the nth term can be obtained as below. Hence Sn can be found. (Where difference of terms are in G.P.) Find sum of n terms of series, 1 + 3 + 7 + 15 + ..... Solution : Sn = 1 + 3 + 7 + 15 + + Tn Sn = 1 + 3 + 7 + +Tn–1 + Tn (Write staggered by 1 place) On subtraction 0 = 1 + 2 + 4 + 8 + + (Tn – Tn–1) – Tn ∴ T = 1 + 2 + 4 + 8 + upto nth term (G.P.) 1(2n − 1) n = 2 − 1 2 − 1 ∴ S = ΣT = Σ2n – Σ1 = (2 + 22 + + 2n) – n 2.(2n − 1) − = 2 − 1 n = 2n+1 – n – 2 Speedy method to Find Tn If the difference between successive terms of a series are in A.P. then its nth term is of the form T and a, b, c can be found by comparison and hence Sn can be found. = an2 + bn + c Find the sum of n terms of the sequence 1, 3, 7, 13, 21, ..... Solution : The differences between successive terms are 2, 4, 6, 8, (in A.P.) Hence, T = an2 + bn + c ∴ T = 1 = a.12 + b.1 + c ⇒ a + b + c = 1 T = 3 = a.4 + b.2 + c ⇒ 4a + 2b + c = 3 T = 7 = a.9 + 3b + c ⇒ 9a + 3b + c = 7 Solving these equations we get a = 1, b = –1 and c = 1 ∴ T = n2 – n + 1 ∴ S = Σn2 – Σn + n = n (2n 2 + 4) = n (n 2 + 2) (on simplification) 6 3 SOLVED EXAMPLES Example 1 : Let a1, a2, a15 be an A.P. such that the A.M. of a1 and a15 is 15. If a7 is given to be 12, find the A.P. Solution : Let ‘d’ denotes the common difference of the A.P. a1 + a15 = 15 2 ⇒ a + a + 14d = 30 ⇒ a + 7d = 15 (1) Also a1 + 6d = 12 (2) Solving (1) and (2) we get, d = 3, a1 = –6 Therefore A.P. is given by –6, –3, 0, 3, 6, 42. Example 2 : The first and the last term of an A.P. having (n + 1) terms are a and b. A new series of n terms is formed by multiplying each of the first n terms by the next consecutive term. Show that the sum of the series is {(n2 –1) (a2 + b2) + (n2 + 2) ab}/3n. Solution : Let d denotes the common difference of the A.P ; then , b = a + nd (1) The new series is given by a (a + d) + (a + d) (a + 2d) + + (a + (n – 1)d) (a + nd) n ⎧⎪ n n ⎫⎪ ⎧⎪ n n ⎫⎪ = ∑(a + (r − 1)d ) (a + rd ) = ∑a2 + ad ⎨2∑r − ∑1⎬ + d 2 ⎨∑r 2 − ∑r ⎬ r=1 r=1 ⎩⎪ r=1 r=1 ⎪⎭ ⎪⎩ r=1 r=1 ⎪⎭ = na2 + a b − a ⎧2n(n + 1) − n⎫ + d 2 ⎧ n (n + 1)(2n + 1) − n(n + 1) ⎫ ⎨ ⎬ ⎨ ⎬ ⎭ ⎩ 2 ⎭ = 2 (b − a)2 n(n + 1) ⎧2n + 1 ⎫ na + a(b − a)n + n2 2 ⎨ 3 − 1⎬ ⎭ = abn + 1 (b − a)2 (n 2 − 1) = 1 (3abn 2 + (a2 + b2 )(n2 − 1) − 2ab (n 2 − 1)) 3n 3n = 1 ((n2 − 1)(a2 + b2 ) + (n2 + 2)ab). 3n Example 3 : If a1, a2, a3, ..., an are in A.P. where ai > 0 for all i, show that 1 + 1 + ..... + 1 = (n −1) an−1 + an a1 + an Solution : L.H.S. = 1 + 1 + ..... + 1 = 1 + 1 + ..... + 1 a2 − = a1 + a3 − a2 + + an − an−1 (a2 − a1) (a3 − a2 ) an − an −1 Let ‘d ’ is the common difference of this A.P. then, a2 – a1 = a3 – a2 = = an – an–1 = d Now L.H.S. = 1 { − + d = 1 { − } d − + .... + − + − } = an − a1 d( + a1 ) = a1 + (n − 1)d − a1 d( = 1 . d = + a1 ) = R.H.S. If a, b, c are in G.P. and log ⎛ 5c ⎞, log ⎛ 3b ⎞ and log ⎛ a ⎞  are in A.P., show that the lengths a, b, c do not ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ form a triangle. Solution : ⎝ a ⎠ ⎝ 5c ⎠ ⎝ 3b ⎠ Given b2 = ac and 2log ⎛ 3b ⎞ = log ⎛ 5c ⎞ + log ⎛ a ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Then, ⎛ 3b ⎞2 ⎜ ⎟ 5c = 5c ⋅ ⎝ 5c ⎠ a = 5c ⎝ a ⎠ ⎝ 3b ⎠ ⎝ ⎠ a 3b 3b ⇒ (3b)3 = (5c)3 ⇒ b = 5 b 2 ⇒ c 2 = 25 9 c ac ⇒ c 2 3 = 25 9 ⇒ a = 25 c 9 5 3 Thus sides are b, b, b 3 5 But b + 3 b = 8 b = (1.6)b < 5 b 5 5 3 ⇒ a, b, c can not form sides of a Δ. Example 5 : Sum to n terms the series 0.4 + 0.44 + 0.444 + ............ Solution : Let S denotes the sum to n terms. Then S = 0.4 + 0.44 + 0.444 + upto n terms = 4 {0.9 + 0.99 + 0.999 + upto 9 n terms} 4 ⎧ 9 ⎨ 9 ⎩10 + 99 100 + 999 1000 upto n terms⎫ ⎭ ⎧ ⎛ 1 ⎛ 1 ⎞ ⎞⎫ 4 ⎧⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞  ⎫ 4 ⎪ ⎜ ⎜1− 10 n ⎟ ⎟⎪ = ⎨⎜1− ⎟ + ⎜1− ⎟ + ⎜1− ⎟ + upto n terms⎬ = ⎪ ⎜ ⎝ 10 ⎠ ⎟⎪ 9 ⎩⎝ 10 ⎠ ⎝ 100 ⎠ ⎝ 1000 ⎠ ⎭ 9 ⎨n − ⎜ 1 ⎟⎬ = 4 ⎧⎪ (10n − 1)⎫⎪ ⎪ ⎜ 1− ⎟⎪ ⎩⎪ ⎝ 10 ⎠⎭ 81 ⎨⎪9n − 10n ⎬ . ❑ ❑ ❑