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PROGRESSIONS 3. l)etermine k, so that k+2, 4k6 and 3k2 are three consecutive terms of an AR (a) 5 (h) 7 (c) 9
Arithmetic Progression (A.P.) (ii) Sum of the n terms s, A sequence whose terms increase 1 or decrease by a fixed number is = a I TI !
i (d) In an A.P., we usually denote the (in) If Irk I the sum of infinite G.P. Solution irst term by 'a', the common differ
k+2 4k 6 3k 2 are in
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a) SItS h) 512 504 510 Solution: Let 'a be the first term and r'the common ratio
1±131 6
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(4k 6) (k+2) = (3k 2) (4k 6) But n ~ ckarl', d = t iii n= 22 a :Scc=.13 .>3k8=k+4 r nth term t = a + (n l)d The 4k = 12 k=3 6. Find the sum of the first hundred Results of G p Sum of ii terms 4 Determine the sum of the first 35 men nalural numbers divisible h (1) If all terms of a G.P. be multi terms of an A.P S =.]2a+(n l)dj a) plied or (litided h the same non If t2 = 2 and t7 = 22 b) 50560 zero constant, then it remains a (a) 2,510 = (a+!). where 1 is the last c) 50500 G.P. will the same common ratio. d) 50505 (2) The reciprocal of the terms of a (b) 2310 term. (c) 2710 given G.P. forms a G.P. Solution: Even natural numbers (3) If each term of a G.P. be raised (d) 2910 Results of AP which are ditisible by 5 are II) 2(1 to same po"ser, the resulting j Solution Let a and d be the first 30, 40... (i) If the same number is added or term and the common difference They form an A? cith a = 10 d = 10 sequence also forms G.P. subtracted to each term of A.P., the resultant sequence is also an i 4 Any 3 consecutive terms of a G.P. respectively. = t = a+('s 1) d A P havi ng some common differ Sil. 10(1 [2x10+(l(9) I) x 10] can be taken as " a ar an 4 con 2 a+d=2 (i) : enee. t, = a+(21)d=2 : (2) If each term of A.P. is multiplied secutise te rms of a G.P. can be and t = a + (7 l)d = 22 .= a+bd =  o 20+ 990  O 1 01 0 50500 or divided by the same number taken as 22 (a) ar ar' and so on 11 h L' ' h r 4. : 1 in the sum o a integers etimeen : W. The IC result•ing sqeuence is lso : Subtracting (i) from (a), we get 50 and 500 which are di',isible by? If = 4 ac. b c are in G . then b2 an A.P. : : Sd = 20 d=4 :a) 1966 9 In a fi nite A.P, the sum of the :GeometricMean(G.M):lfa.x,bare: tUSifl. a 9 b) 1177996 terms equidistant from the in G.P, thenx = J55 is the G.M of a t.I) c) 17766 beginning and end is always same and b. , [2a+ (nI) dJ :Now,S =: . n 2 : d) 17696 and equal to the sum of fi rst and : Multi ple Choice Questions: : Solution: The fi rst integer, after .50 last term. S35 =115 [4+(35l) 4] 1. Determine 25th term of an A.P. 2 which is dMsihle by? is 56 and the 6', 14 whose 9th term is 6 and common last intcger before 500 which is =4 X 132 = 35 x 66 = 2310 246 8 10 12 difference is divisible by 'is 49. 5. How many terms of the A. P. 14 7 I (a) 16 The sequence of mtegers between are needed to give the sum 715? 14 (b) 18 :o0andaO0whicharedivisiblebyi is :(a.,.3.3 If a, b, care in A.P., (c) 12 56, 63, 70 'b" then2b=a+c' i1tisanA.P.witha=56,d=7 1(d) 14 i (5) The common difference can be Solution Let a be the fi rst term a = 497 = a + (n  1) d tdi 2 zero, positive or negative, :• and 'd' the common difference of an 497 = 56 + (n l)X 7 Solu tion: Here a=1, d=3. :. (6). If a' is the fi rst term and 'd' the : A P  '*91 ' I r)O Let 715 be (lie sum of n terms of common difference of ii A P .Then a,, = a+ (nI) d : or in = 448 this A.P. : ' havi ng in terms, then n term : 01', n = 448 '7 = 64. as = a+ (9l)(.) from the end = a+(mn)d is, S9 = 715
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10) What term of pprogression :18
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c) 9 : :d) 12 12 Solution: here a =l8,r= = 18 : : Let 729be the n" term 512 so that a = 729
I S (') I>. It 4th a n d Sib :t rats of a G P are 24 and 384 respectielv then find out first term and common ratio: (a) 2,3 (h) 5, 3 (c) 3,2 (d) None of these Solution: Let 'a' be the first term and r the common ratio Then, :. 4th term = 24 >ar=24......(i) and8thterni=384 ar =384 (n) Dividin" GD by (D we get = .
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Substituting r=2 in (i), we get a= at ) = Hence, fi rst term=3 and : ,ommon ratio : 13. Th e fi rst term of a G,P. is 1. The sum of the third and fifth terms is 90 Find the common ratio of the G.P. Since a9 = ar" (b) :3,3 (a) 2, 3 l2  , F1119 (d)o,3 (c) 13 L 3 j :  729 Solution: Let 'r' be the common 512 256 ratio ofG.P. ' [] 79><5 9x729 First term a=I F1  _11_ Now , aar2r2  aJ L : 3  32x .36 and a =arl=ri But a1ea =90 n1 =8 n=9 r+r'=90 'r1+r9O=Q : Hence jis the 9th term of (+lO) (9)=o : :
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=114(56+497) Putting values, of and d a+b a+b/2 is the AM. of : 2 : : a and b = 16+30=14 [2x1+(nl)x3j = 715 = 32x(553) = 17696 Geometric Progression ((..P.) I 12. Which term of the A.R 5, 1:3,21 I A sequence ( fi nittor infinit:)of 8 Vmrite down the 20th (cmi of the is 181 nonzero numbers in which even  (2+3n3) = 719 G P 1 1 1 1 term except the first one bcars a (a) 21st a) 1 constant ratio with its preceding (b) 22nd 2 3nn 1430 = 0 1 b) 1 term is called a geometric pro (c) 23rd c) + I (d) 24th 1±Jl4 (3) (14 30) gn ssion abriviated as G.P. .. n= 1 d) None of these The ratio of any term ofa G.P. to the I Solution: Here, first term a = 5 2x3 1 Solution: Here, a = I, r= 1 preceding term is called the corn I Common difference d = 8 : Since a,, = ar , mon ratio of the G.P. and is denot : Let 181 be the nth i .e. an = 181 : r r clearls ed b 181=9 ~ (n 1) 8 or 176 = (ii 1) 8 '+bx+c=tl a = am' = lx (_1)' = 1 = thl .. n1 = 176  8=22 x=h±'h2 4ae I th (i) Then term t = n= 23 2a9) The 9th term of a ( P is 2 find ar' Hence, 181 is the 2:3rd term. I the product of first 9 terms.
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19=0 progression ( r+10#0) .'.r=±:3 liThe 3rd term of a G.P. is thel square of the first term. If the 14. In a (IF, the fi rst te rm is 7, the last term 448 and the sum 889. second term is 8 determine the Find the common ratio 6th term. (a) 4 (b) 6 I (b) 1:32 (a) 136 (d) 2 (c) 8 (c) 18 (d) 124 Solution Here a=7 l=a, =448 Solution Let a be the first 'nterm and 'r' be the common Let r be the common ratio ratio of G.P. a(1r ) = we hae a 1=(a1 ) ar=a Sn 1r lr r2=a— I 88 ,9= 748r : Alsoa,=8=ar=8—(n) lr : ' \lultiplsing (I) and (ii) we get 88989r= i448r ar8xa .. r=t3 'r=2 i 889i=889r448r I'rom (i) =(9) =4 [ a=r 882 = 441r r=2 Hence, a,,=ar5=(4) ,'
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