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https://docs.google.com/document/d/1NV-75kykcYsnzTHiZDg9N7wEPOIpsHzL/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true PROGRESSION AND SERIES SEQUENCE A succession of numbers formed according to some definite rule is called a sequence. For example 1, 3, 5, 7,9 ……. is a sequence, here each term of the sequence can be obtained by adding 2 to the preceding term. Types of Sequence There are two types of sequence. i) Finite sequence ii) Infinite sequence a sequence is said to be a finite or infinite according as it has finite or infinite number of terms. Series If {fn} be a sequence then an expression of the form f1 + f2 + …… + fn is called series. In other word a series is the sum of the terms of the sequence. PROGRESSION If the terms of a sequence are written under specific condition then the sequence is called progression. There are three types of progressions. i) Arithmetic Progression ii) Geometric Progression iii) Harmonic Progression ARITHMETIC PROGRESSION (A. P.) An A.P. is a sequence whose terms increase or decrease by a fixed number, called the common difference of the A.P. nth Term and Sum of n Terms: If a is the first term and d the common difference, the A.P. can be written as a, a + d, a + 2d, .... The nth term an is given by an = a + (n - 1)d. The sum Sn of the first n terms of such an A.P. is given by (a + l ) where l is the last term (i.e. the nth term of the A.P.). Notes If a fixed number is added (subtracted) to each term of a given A.P. then the resulting sequence is also an A.P. with the same common difference as that of the given A.P. If each term of an A.P. is multiplied by a fixed number(say k) (or divided by a non-zero fixed number), the resulting sequence is also an A.P. with the common difference multiplied by k. If a1, a2, a3.....and b1, b2, b3...are two A.P.’s with common differences d and d′ respectively then a1+b1, a2+b2, a3+b3,...is also an A.P. with common difference d+d′ If we have to take three terms in an A.P., it is convenient to take them as a - d, a, a + d. In general, we take a - rd, a - (r - 1)d,......a - d, a, a + d,.......a + rd in case we have to take (2r + 1) terms in an A.P If we have to take four terms, we take a – 3d, a – d, a + d, a + 3d. In general, we take a – (2r – 1)d, a – (2r – 3)d,....a – d, a + d,.....a + (2r – 1)d, in case we have to take 2r terms in an A.P. If a1, a2, a3, ……. an are in A.P. then a1 + an = a2 + an-1 = a3 + an –2 = . . . . . and so on. Arithmetic Mean(s): If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a,b,c are in A.P. then is the A.M. of a and c. If a1, a2, ...an are n numbers then the arithmetic mean (A) of these numbers is The n numbers A1, A2......An are said to be A.M’s between the numbers a and b if a, A1, A2,........An,b are in A.P. If d is the common difference of this A.P. then b = a + (n + 2 - 1)d ⇒ ⇒ . Illustration 1: If the Ist and the 2nd terms of an A.P are 1 and –3 respectively, find the nth term and the sum of the Ist n terms. Solution: Ist term = a, 2nd term = a + d where a = 1, a + d = -3, ⇒ d = -4 (Common difference of A.P.) we have an = a + (n –1)d = 1 + (n – 1) (-4) = 5 – 4n Sn = {a + an} = {1 + 5 – 4n} = n (3 – 2n) Illustration 2: If 6 arithmetic means are inserted between 1 and 9/2, find the 4th arithmetic mean. Solution: Let a1, a2, a3, a4, a5, a6 be six arithmetic means . Then 1, a1, a2, …, a6, will be in A.P. Now, = 1 + 7d ⇒ = 7d ⇒ d = Hence a4 = 1 + 4 = 3 Illustration 3: Prove that the numbers , , cannot be any three terms (not necessarily consecutive) of an A.P. Solution : Let , , are the nth, mth and pth term of an A.P. whose first term is a and common difference d ⇒ = a + (n –1)d ; = a + (m –1)d; = a + (p –1)d; so ; ∴ irrational = rational. Illustration 4: The interior angles of a polygon are in arithmetic progression. The smallest angle is 120o and the common difference is 5o. Find the number of sides of the polygon. Solution: Let the number of sides of the polygon be n. The sum of the interior angles of the polygon = ( n -2) π = (n-2).180o Also the first term of the A.P.= a = 120o The common difference = d = 5o For n=16, the largest angle = 120o+15 x 5o = 195 o. This is not possible as an interior angle of a polygon cannot be greater than 180o. Hence n = 9. Illustration 5: Let Sn denote the sum upto n terms of an A.P..If Sn = n2P and Sm = m2P , where m, n and p are positive integers and m ≠ n, then find Sp. Solution: Let first term be a and common difference be d, then Sn = (2a + (n –1)d) = n2P Sm = (2a + (m –1)d) = m2P ⇒ (n –m)d = (2n –2m)P ⇒ d = 2P and a = p ⇒ Sp = [2a + (p –1)d] = [2p + (p –1)2p] ⇒ Sp = p3. GEOMETRIC PROGRESSION (G.P.) A G.P. is a sequence whose first term is non-zero and each of whose succeeding term is r times the preceding term, where r is some fixed non - zero number, known as the common ratio of the G.P. For example 3 + 9 + 27 + 81 is a G.P. whose common ratio is 3. nth Term and Sum of n Terms: If a is the first term and r the common ratio, then G.P. can be written as a, ar, ar2, . . . the nth term, an, is given by an = arn-1. The sum Sn of the first n terms of the G.P. is If -1 < r < 1, then the sum of the infinite G.P. a + ar + ar2 +........= Notes If each term of a G.P. is multiplied (divided) by a fixed non-zero constant, then the resulting sequence is also a G.P. with same ratio as that of the given G.P. . If each term of a G.P. (with common ratio r) is raised to the power k, then the resulting sequence is also a G.P. with common ratio rk. If a1, a2, a3, .... and b1, b2, b3, .... are two G.P.’s with common ratios r and r′ respectively then the sequence a1b1 , a2b2, a3b3.....is also a G.P. with common ratio r r′. If we have to take three terms in a G.P., it is convenient to take them as a/r, a, ar. In general, we take in case we have to take (2k + 1) terms in a G.P. If we have to take four terms in a G.P., it is convenient to take them as a/r3, a/r, ar, ar3 . In general, we take , in case we have to take 2k terms in a G.P. If a1, a2, …., an are in G.P., then a1an = a2 an-1 = a3 an-2 = . . . . If a1, a2, a3,.…is a G.P. (each aI > 0), then loga1, loga2, loga3 ….. is an A.P. The converse is also true. Geometric Means If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P. then b = is the geometric mean of a and c. If a1, a2......an are non-zero positive numbers then their G.M (G) is given by G = (a1a2a3......an)1/n. If G1, G2,…..Gn are n geometric means between a and b then a, G1, G2, …., Gn, b will be a G.P. Here b = a rn + 1 ⇒ r = ⇒ G1 = a , G2 = a, . . . . . . , Gn = a Illustration 6: The third term of a G.P. is 7. Find the product of first five terms. Solution: Let the terms be , a , ar, ar2 ⇒ a = 7. The product = a5 = 75. Illustration 7: If ax = by = cz and x, y, z are in G.P., show that logb a = logc b Solution: Given, ax = by = cz = k (say) Taking logarithm, we get x log a = y log b = z log c = log k ∴ x = , y = and z = since x, y, z are in GP. ∴ ∴ × = × ⇒ = ⇒ logba = logcb Illustration 8: The sum of three numbers in GP is 21 and the sum of their squares is 189, find the numbers. Solution: Let the three numbers be a, ar, ar2 Given a + ar + ar2 = 21 ⇒ a (1 + r + r2) = 21 ...(1) And a2 + a2r2 + a2r4 = 189 ⇒ a2( 1 + r2 + r4) = 189 …(2) Squaring (1) and dividing (2) by it, we get ⇒ ⇒ ⇒ ⇒ 7 ( 1+ r2-r ) = 3( 1+ r+ r2) ⇒ 4r2 –10r + 4 = 0 ⇒ 2r2 –5r + 2 = 0 ⇒ r = 2, Hence the three numbers are 3, 6, 12 or 12, 6, 2 Illustration 9: Find the sum of upto infinite terms. Solution: ARITHMETICO-GEOMETRIC PROGRESSION Suppose a1, a2, a3......is an A.P. and b1, b2, b3..... is a G.P. Then the Progression a1b1, a2b2, ....... is said to be an arithmetico-geometric progression (A.G.P). Hence an arithmetico-geometric progression is of the form ab, (a+d)br, (a+2d)br2, (a+3d)br3,......... Sum of n Terms: The sum Sn of first n terms of an A.G. P. is obtained in the following way : Sn = ab + (a + d)br + (a + 2d)br2 +.........+(a + (n - 2)d)brn-2 + (a + (n - 1)d)brn-1 Multiply both sides by r, so that r Sn = abr + (a + d)br2+.........+(a + (n - 3)d)brn-2 + (a + (n - 2)d)brn-1 + (a + (n - 1)d)brn Subtracting, we get (1 - r)Sn = ab + dbr + dbr2 +.......+dbrn - 2 + dbrn - 1 - (a + (n - 1)d)brn = ⇒ If -1 < r < 1, the sum of the infinite number of terms of the progression is = . Illustration 10: Find the sum of the series 1.2 + 2.22 + 3.23 + ….. + 100.2100. Solution: S = 1.2 + 2.22 + 3.23 + …. + 100.2100 2S =1.22 + 2.23 + …. + 99.2100 + 100.2101 ⇒ –S = 1.2 + 1.22 + 1.23 + …. + 1.2100 – 100.2101 ⇒ -S = 1.2 - 100.2101 ⇒ S = -2101 + 2 + 100.2101 = 99.2101 + 2. Illustration 11: Find 1 + 2.2 + 3.22 + 4. 23 + …….+ 100.299. Solution: Let Sn = 1 + 2.2 + 3.22 + ……+ tn ……(1) then 2Sn = 1.2 + 2.22 + 3.23 + ……+ 2.tn –1 + 2. tn ……(2) Subtracting (2) from (1), we get -Sn = 1 + [(2 –1) 2 + (3 –2) 22 + …to (n –1) terms] –2. tn = 1 + 2 + 22 + ….to n terms – 2. tn = …..(3) Now tn = (nth term of AP 1, 2, 3, 4, ….) . (nth term of GP 1, 2, 22, …) = {1 + (n –1) .1} {1. 2n –1 } = n2n –1 from (3), Sn = -(2n –1) + 2n2n –1 = 1 –2n –n2n = 1- (1 + n)2n HARMONIC PROGRESSION (H.P.) The sequence a1, a2, a3.......an......(ai ≠ 0) is said to be an H.P. if the sequence is an A.P. nth Term of H. P.: The nth term, an, of the H.P. is Note There is no formula for the sum of n terms of an H.P. Harmonic Means If a and b are two non-zero numbers, then the harmonic mean of a and b is a number H such that the numbers a, H, b are in H.P. We have If a1, a2, .......an are ‘n’ non-zero numbers, then the harmonic mean H of these numbers is given by . The n numbers H1, H2,.......,Hn are said to be n-harmonic means between a and b, if a , H1 , H2 ........, Hn , b are in H.P. i.e if are in A.P. Let d be the common difference of the A.P., then ⇒ d = Thus Illustration 12: Find the 4th and the 8th terms of the H.P. 6, 4, 3,………. Solution: Consider Here T2 – T1 = T3 – T2 = ⇒ ,….. is an A.P. 4th term of this A.P. = + 3 × = + = , and the 8th term = + 7 × = Hence the 4th term of the H.P. = and the 8th term = Illustration 13: x + y + z = 15 if a, x, y, z, b are in A.P. and if a, x, y, z b are in H.P. Find a and b. Solution: Given x + y + z = 15 ..(1) when a, x, y, z, b are in AP. ∴ sum of A.M.’s, x + y + z = ∴ 15 = .3 ⇒ a + b = 10 …(2) when a, x, y, z, b are in HP., are in AP. ∴ .3 ⇒ ∴ ab = 9 ..(3) Now (a –b)2 + (a + b)2 –4ab = 100 –36 = 64 ∴ a – b = ±8 …(4) from (1) and (4) , we get a = 9, b = 1 or a = 1, b = 9 MISCELLANEOUS PROGRESSIONS Some Important Results: 1 + 2 + 3+…+n = (sum of the first n natural numbers) 12 + 22 + 32 + … + n2 = (sum of squares of the first n natural numbers) 13+23+33+…+n3==(1+2+3+…+n)2 (sum of cubes of first n natural numbers) 1 + x + x2 + x3 +........ = (1 - x)-1, if -1 < x < 1 1 + 2x + 3x2 +........... = (1 - x)-2, if -1 < x < 1 Method of Differences Suppose a1, a2, a3, .......is a sequence such that the sequence a2 - a1, a3 - a2, .........is either an A.P. or a G.P. The n th term ‘an’ of this sequence is obtained as follows : S = a1 + a2 + a3 +........+an-1 + an S = a1 + a2 +.........+an-2 + an-1 + an ⇒ an = a1 + [(a2 - a1) + (a3 - a2) +......+(an - an - 1)] Since the terms within the brackets are either in an A.P. or in a G.P., we can find the value of an, the nth term. We can now find the sum of the n terms of the sequence as Illustration 14: Find the sum of n terms of the series 1+ 5 + 11 + 19 + 29+ ………. Solution: Here the difference of consecutive terms of the series are 4, 6, 8, 10… which are in AP. Let Sn = 1 + 5 + 11 + 19 + ……+ tn –1 + tn …(1) And Sn = 1 + 5 + 11 + 19 + ……+ tn –1 + tn …(2) Subtracting (2) from (1), we get 0 = 1 + {4 + 6 + 8 + …..to (n –1) terms} –tn or tn = 1 + {4 + 6 + 8 + …..to (n –1) terms} = 1 + {2.4 + (n –1 –1)2} = 1 + (n –1) (n + 2) = 1 + n2 +n –2 = n2 + n – 1 ∴ Sn = ∑ tn = ∑ (n2 + n –1) = ∑ n2 + ∑ n - ∑ 1 = = = Illustration 15: Find the sum of Ist n terms of the series 5, 7, 11, 17, 25,…….. Solution: Let S = 5 + 7 + 11 + 17 + 25+……….+ tr S = 5 + 7 + 11 + 17 + ….…… + tr – 1 + tr Subtracting, we get 0 = 5 + 2 + 4 + 6 + 8 + ……..+ rth term – tr ⇒ tr = 5 + 2 ⇒ tr = r2 – r + 5 Sn = = + 5n = {(n + 1)(2n + 1) – 3(n + 1) + 30} = (n2 + 14). Illustration 16: Find the nth term and the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 + ……… Solution: rth term of the series = r.(r + 1).(r + 3) = r3 + 4r2 + 3r Hence the sum of n terms = + 4 + 3 = = {3n2 + 19n + 26} INEQUALITIES A.M. ≥ G. M. ≥ H. M.: Let a1, a2,... . . . , an be n positive real numbers, then it can be shown that A ≥ G ≥ H. Moreover equality holds at either place if and only if a1 = a2 = ……. = an . Weighted Means: Let a1, a2, a3 . . . , an be n positive real numbers and m1, m2, . .. . . . , mn be n positive rational numbers. Then we define weighted Arithmetic mean (A*), weighted Geometric mean (G*) and weighted harmonic mean (H*) as A* = , G* = and H* = . It can be shown that A* ≥ G* ≥ H*. Moreover equality holds at either place if and only if a1 = a2 = . . . . . = an . Arithmetic Mean of mth Power: Let a1, a2,......, an be n positive real numbers (not all equal) and let m be a real number, then > if m ∈ R –[0, 1]. However if m∈ (0, 1) , then < . Obviously if m∈ {0, 1} , then = . Illustration 17: Show that x +≥ 2, if x > 0 and x + ≤ -2 , if x < 0 . Solution: Since x > 0, ≥ ( A.M. ≥ G. M. ) ⇒ x+≥ 2. If x < 0 , let y = -x , then y > 0 and y + ≥ 2 ⇒ -x - ≥ 2 ⇒ x + ≤ -2. Illustration 18: If ai ’s are all positive real numbers then prove that (1+ a1 +a12 ) (1+ a2 +a22 ) . . . . (1+ an +an2 ) ≥ 3n a1a2 ….. an . Solution: ≥ (1. ai . ai2 )1/3 = ai ( i = 1, 2, 3, . . . . , n) ⇒ 1+ai +ai2 ≥ 3ai ⇒ (1+ a1 +a12) (1+ a2 +a22) . . . . (1+ an +an2 ) ≥ 3n a1a2 ….. an . Illustration 19: If x, y, z are positive real numbers such that x + y + z = a, then prove that . Solution: Since A.M. ≥ H.M. ⇒ . Illustration 20: If a, b, c are positive real numbers such that a + b + c = 1, then prove that . Solution: We have to show that i.e Now A.M. of mth power ≥ mth power of arithmetic mean ( m = -1 here) ⇒ ⇒ . ASSIGNMENT 1. The sum of the series (n).1 + (n –1).2 + (n –2).3 + ……+ 1.n is (A) (B) (C) (D) 1. The rth term of the series is given by Tr = (n –r + 1)r Sum of the series = ⇒ Sn = ⇒ Sn= = Hence (A) is the correct answer. 2. If a, b, c be positive real numbers forming a H.P., then is always equal to (A) 2/a (B) 2/b (C)2/c (D) None of these 2. (B) 3. If a1, a2, a3, ..., an are in H. P, then a1a2 + a2a3 + a3a4 + …+ an-1 an is equal to (A) (n -1) a1an (B) na1an (C) (D) none of these. 3. a1, a2 , a3 , . .. . . . , an are in H. P. Then are in A.P. ⇒ = d (say) . . . . (A) ⇒ a1a2 + a2a3 + a3a4 + . . . . + an-1an = (a1 -an) . .. . . (B) If we add all (n-1) terms of (A), we get d ⇒ . Thus from (B) a1a2 + a2a3 + a3a4 +. . .. . . . + an-1an = ( n-1)a1an . Hence (A) is the correct answer. 4. If be the A.M. and be two G.Ms between two positive numbers a and b, then is equal to (A) 2 (B) 1 (C) 2 (D) None of these 4. (C) 5. If the sum to n terms of a series be 5n2 + 2n, then second term is (A) 15 (B) 17 (C) 10 (D) 5 5. Sn = 5n2 + 2n, Sn –1 = 5(n –1)2 + 2(n –1) ⇒ Tn = Sn –Sn –1 = 10n –3 ⇒ T2 = 20 –3 = 17 Hence (B) is the correct answer. 6. of a triangle ABC are in A.P. If a, b, c are the corresponding sides, then (A) (B) (C) (D) None of these 6. (B) 2 7. The sum of the n terms of two A.P series are in the ratio of (n + 1) : (n + 3). The ratio of their 4th terms is (A) 3/4 (B) 4/5 (C) 1/2 (D) 1/3 7. Let Sn and be sum of n terms of two series then ⇒ a1 = 1, d1 = 1, a2 = 2, d2 = 1 ⇒ . Hence (B) is the correct answer. 8. Sides of a triangle ABC; a, b, c are in G.P. If ‘r’ be the common ratio of this G.P., then (A) (B) (C) (D) 8. (C) Let b = ar, c = a For triangle ABC, a + b > c a + ar > a - r – 1 < 0 < r < Also, b + c > a a (r + ) > a + r – 1 > 0 r Finally, a + c > b a [1 + ) > ar - r + 1 > 0. which is true for all values of r. In this case ‘r’ can not be negative r 9. If a, b, c are in G.P then loga x, logb x, logc x are in (A) A.P (B) G.P (C) H.P (D) none of these 9. Given that a, b, c are in G.P ⇒ b2 = ac ⇒ 2 logx b = logx a + logx c ⇒ ⇒ loga x, logb x, logc x are in H.P. Hence (C) is the correct answer. 10. If a,b,c,d and a, b, c d are in H.P., then (A) a + d > b + c (B) a + b > c + d (C) a + c > b + d (D) None of these 10. (A) Using A.M. H.M. we get > b, a + c > 2b and > c, b + d > 2c a + c + b + d > 2b + 2c a + d > b + c 11. The sum of the series 1 + 4 + 3 + 6 + 5 + 8 + ……. upto n terms when n is an even number is (A) (B) (C) (D) 11. The series can be written as S = (1 + 3 + 5 + 7 + …..n/2 terms) + (4 + 6 + 8 + 10 + …..n/2 terms) S = = [n] + [8 + n –2] = = Hence (B) is the correct answer. 12. The determinant Δ = is equal to zero, if (A) a, b, c are in A.P. (B) a, b, c are in G. P. (C) a, b, c are in H.P. (D) α is a root of ax2+2bx+c=0 12. It is easy to see that Δ = (b2 –ac)(aα2 +2bα +c) . Hence Δ =0 if a, b, c are in G.P. or α is a root of ax2 +2bx +c = 0. Hence (B) and (D) are the correct answers. 13. The sum of the first n terms of the series 12+2.22+32+2.42+52+2.62+ . . . is , when n is even. When n is odd, the sum is (A) (B) (C) (D) 13. If n is odd, n-1 is even. Sum of (n-1) terms will be . The nth term will be n2. Hence the required sum = +n2 = Hence (A) is the correct answer. 14. , i = 1, 2, ….n is arithmetic sequence. If then is equal to:[ (A) 600 (B) 900 (C) 300 (D) None of these 14. (B) = 6 + (4d + 9d + 14d + 19d + 23d) = 3 (2 + 23d) = 225 (given) Now, (2 + 23d) = 12 75 = 900 15. Given p A.P's, each of which consists of n terms. If their first terms are 1, 2, 3, …, p and common differences are 1, 3, 5, ---, 2p –1 respectively, then sum of the terms of all the progressions is (A) np(np+1) (B) n(p+1) (C) np(n+1) (D) none of these . 15. The rth A. P. has first term r and common difference 2r – 1. Hence sum of its n terms = . The required sum = = = = . Hence (A) is the correct answer. 16. , i = 1, 2, ….n is arithmetic sequence. If then value of common difference of the A.P. such that is minimum, is equal to (A) (B) (C) (D) None of these 16. (C) = = = (say) (.5 + 5 + 9) = 15 Clearly, is the point of minima for f() d = 17. The third term of a G.P. is 4, the product of the first five terms is (A) 43 (B) 45 (C) 44 (D) none of these 17. Let the first five terms of the given G.P. be a1 , a2 , a3 , a4, a5 . Hence a3 = 4 . Now a1 a5 = a2 a4 = a32 ⇒ a1 a2 a3 a4 a5 = 45. Hence (B) is the correct answer. 18. If the product of n positive numbers is unity , then their sum is (A) a positive integer (B) divisible by n (C) equal to n +1/n (D) never less than n. 18. Let the numbers be a1, a2, a3, . . ., an. Then a1.a2.a3 . .….an =1. Using A.M. ≥ G.M , we get ⇒ a1 + a2 + a3 + . . . . + an ≥ n Hence (D) is the correct answer. 19. If a, b, c are in H.P. , then the value of is (A) 0 (B) 1 (C) 2 (D) 3 19. a, b, c are in H.P. ⇒ b = ⇒ ⇒ . . . . (A) Again a, b, c are in H.P. ⇒ b = ⇒ ⇒ . . . . (B) From (A) and (B) =2. Hence (C) is the correct answer. 20. If first and (2n-1)th terms of an A.P. , G. P. and H.P. , are equal and their nth terms are a, b, c respectively , then (A) a+c = 2b (B) a+c = b (C) a ≥ b ≥ c (D) ac –b2 = 0 20. Let α be the first and β be the (2n-1) terns of an A.P. , G.P. and H.P. , then α, a, β will be in A.P. , α, b, β will be G.P. α, c, β will be in H.P. Hence a, b, c are respectively A. M. , G.M. and H.M. of α and β. Since A.M. ≥ G.M. ≥H.M. , a ≥ b ≥ c. Again a = , b2 = αβ and c = . Hence ac-b2 =0. Hence (C) and (D) are correct answers. 21. If a, b and c are positive real numbers then is greater than or equal to (A) 3 (B) 6 (C) 27 (D) none of those 21. Using A. M. ≥ G. M. ⇒ ≥ 3. Hence (A) is the correct answer. 22. If a, b, c are in A.P., then value of the expression is also equal to (A) 2abc (B) 6abc (C) 4abc (D) None of these 22. (D) 2b = a + c 8= = + + 3ac (a + c) 8= + + 3ac(2b) + - 8= -6 abc 23. The sum of the infinite series . . . . is: (A) (B) (C) (D) 23. (A) S = … ∞ = = 24. If a, b, c are in H.P., then is always equal to (A) (B) (C) (D) 24. (B) 25. The sum of n terms of the series is: (A) (B) (C) (D) None of these 26. The sum of infinitely many terms of the series is: (A) 0 (B) 4 (C) 6 (D) Can't be determined 25/26. Sn = = = 6 = Ans. of 3 question Now, S∞ = Sn = = = 6 Ans. of 4 question 27. If H be the harmonic mean of a and b then the value of is: (A) 0 (B) 2 (C) 1 (D) a + b 27. (A) ⇒ − 2 = 0 28. The p th term of an A.P. is a and q th term is b. The sum of its (p + q) terms is : (A) (B) (C) (D) 28. (A) a = a1 + (p − 1) d … (1) b = a1 + (q − 1) d … (2) where a1 is the first term and d is the common difference of the A.P. Now, a − b = (p − q) d ⇒ d = and a + b = 2 a1 + (p + q − 2) d = 2 a1 + (p + q − 1) d − d … (3) Sp+q = [2 a1 + (p + q − 1) d] = [a + b + d] = 29. If term of G.P. is ‘a’ and it’s terms is ‘b’ where a, b then it’s term is:[ (A) (B) (C) (D) None of these 29. (C) Let ‘A’ be first term and ‘r’ be the common ratio. Then a = term 30. The sum of the series to infinity: is: (A) (B) (C) (D) 30. (C) S = + … ∞ = =