5 a Geometric Progre: A list of numb sion (G sin which each term is by a fixed (non-zero) number, except the firs (abbreviated G.P). i 2 erm to its precedin, Thus, a list of numbers forms a G.P. if and only if the rae of any ter P 8 term is constant (a fixed non-zero number), tiplying, its preceding term ned by mul obtained PY , geometric progression st term, is called 2 This fixed number is denoted by r. The first, second, third, .... terms of a GP. are respectively So, the list of numbers a), dy, ay, ... forms a G.P. if and only if ca The first term of a GP. is also denoted by 4. The general G.P. is a, ar, ar, «0 GP. is called finite if and only if it contains a finite If ay, ay ay... is a GP, then a, + ay + ay + --- is calles General term of a G.P. If a, ar, ar’, .... is a GP, then its general term is denoted by Ifa finite GP. a ar, ar... contains 1 term and its last term is | then The nth term from end of a finite G.P. denoted by 4, 4 4% 1, a fixed number. number of terms. .d a geometric series. a, and a, = a7" ~*. Dear) If a, ar, ar, ... is a finite G.P. with last term J, then nth term from end = 1 ( ) Numbers in G.P. 1. The numbers a, b, ¢ are in GP. if and only if = Sie. iff b? = ac. 2. If the product of numbers in G.P. is given, then (i three numbers are taken as ©, , ar. (if) four numbers are taken as ©, ©, ar, ar°. ly (iii) five numbers are taken as 4,4, a, ar, ar. Pr Sum of first n terms of a G. Ifa is the first term and r is the common ratio of a GP. and the sum of its first n terms is denoted by S,, then _ adler") = ar" =1) @ 8,= G2 ors, = SSP ret. Note that if r= 1, then S,=a+a+a+... 1 times = na. (ii) If Lis the last term of G.P, then 5, = 1 or, =8.2 GEOMETRIC PROGRESSION (6.P) A list of (non-zero) numbers is called a geometric progression (abl Ff the ratio of its any term to its preceding term is constant i.e. a fixed mumber: This (non-zero) constant (a fixed number) is usually denoted by r and is called common ratio. Note that this fixed number r may be positive or negative. The first, second, third, terms of a G.P. are respectively denoted by dy, Ay, yy v++ OF thy tay fay o-+ Thus, a, ay ay... is a GP. if and only if 4 = r, a constant (independent of i) i.e. breviated G.P) if and only a, r (except the first term). The first term of a GP. is also denoted by a. It follows that ina GP, a, ,, = 4, r ie. any term (except the first) is obtained by multiplying its preceding term by a fixed (non-zero) number r. For example Consider the following lists of numbers: | (i) 2, 4, 8, 16, 32, ..., 1024 9.24 Letaon Iso 8? 19) ire he 8) he ble nly on Sieeisy iauisliemil (iii) (o) 4,-1,3,-9. In list (9, each term is obtained by multiplying its preceding term by 2 (except the first term), so it is a GP. In list (i, each term is obtained by multiplying its preceding term by (~2] (except the first term), so it is a GP. oe » a) er In list (ii), each term is obtained by multiplying its preceding term by 5 (except the first term), so it isa GP, In list (iv), each term is obtained by multiplying its preceding term by (- 3) (except the first term), so it is a G A geometric progression is called a finite GP. if and only if it contains a finite number of terms; and a G.P.is called infinite if and only if it contains an infinite number of terms. ‘Thus, in the above example, each of (i) and (fi) is a finite G.P; each of (fi) and (fo) is an infinite GP. If a GR. is given, then its first term and common ratio can be found. For example: ( HG.P.is2,4,8,16, 32... 1024, then its first term a=2 and common ratio= Note that r can be obtained by using any two consecutive terms. then its first term a = 4 and common ratio = “2 cy IGPis4,-2,1,-3,. Conversely, if the first term and the common ratio of a GP, are known then the GBP. can be obtained. If a be the first term and r be the common ratio, then the GP. is a, ar, a, a7, For example: (i If the first term and the common ratio of a GP. are 5 and 2 respectively ie. a = 5 and r= 2, then the GP. is 5, 5x25x2, es ie. 5, 10, 20, 40, ... (ii) If the first term and the common ratio of a GP, are 9 and =} respectively ie. y9(-3) 4, then Gis 9, 9 x (-4).9* (- Tay ty, dy se ate in GP. then a, + 45 + 05 +... +a, is called a geometric series. 9.2.1 General term of a G.P. Let a be the first term and r(¢ 0) be the common ratio of a GP. Since, any term (except the first) is obtained by multiplying its preceding term by r a = ayy = arr = ar = an Pr = ar = ark} ‘Arithmetic and Geometric Progressions| r pattern, we getty at y “ By looking at this then the G.P. is a, ar, ap Hence, general term a, = HF “wofaGh Flee ats the fist germ and ris ee cOMIIEN TTT ye | rel or a, 0 ry erm eeorling aH Hs fiite oF INE ie enoted by I, then i: We that teem of GP. costing of terms Tear" The nth term from the end of a finite GP / (2 | orm from the end i a finite GP with last term /, then the nth term fi ! Wa, ar For, when we look a the terms of the giver GP. from the last and move towards beginning, 1 rst term as I, therefore, wwe find that the Hist of numbers is a GP, with common ratio 2 and fist term a | therefor, nth term from the end of given GP. = (2) 4. Three numbers a,b, are in GR. B= E ie FOP = imbers in GP. | 2. Some problems involve 3, 4 oF 5 nut then ina G.P, If the product of the numbers is given, (0 three numbers are token as © , dar (id four numbers are taken as SA ar, ar. (uh five numbers are taken as 5, a ar, ar 7 Geometric mean If a and b are any two positive real numbers, between a and b. then Jab is called the geometric mean For example: (@ The geometric mean between 4 and 64 = J4x 64 = V256 = 16. (i) The geometric mean between + and + | 3 nd - fee. -2 3a & “> STENTS ety ES oo Example 1, () Find the 20th term and the nth term of the list of numbers 5, 5,5 ms (ii) Find the nth and the 15th term of the series 1- 1,11 ee (iii) Which term of the list of numbers V3, 3, 3/3, i is no : (iv) Which term of the list of numbers 18, ~12, 8, ... is 12> 7 Fr S 3, ~ 6, 12, ~ 24, ..., 3072. HSolution, (i) The is a G.P. with first term a= > and common 1 ratio r = a, =arn1 : 1 (i) The given series is a geometric series with first term a = 1 and common ratio r= ~ 5+ Hence, nth term a, = ar"! = 1x (-y" = (aye 2h, YL (gyi x 215 = 2, In particular, a; = 1 x (- (iii) The given list of numbers is a G.P. with first term a = V3 and common ratio = V3 - . Let 729 be the nth term. Then 4, = 729 = ar'-1=729 = VB (VB) 1 = 3% = (VB = (V3)P a2. Hence, the 12th term of the given list of numbers is 729. (iv) The given list of numbers is a G.P. with first term a = 18 and common ratio r= — Let 32 be the nth term. Then 512 512 of 512 a, = 2 = art = 52 = 18(-2) = 512 | = 729 729 8, 3. 729 Basta (ay | 18° 729° 9x79 3 13 => n-1=8 >n= 7 | Hence, the 9th term of the given list of numbers is =>. (2) The given list of numbers 3, - 6, 12, - 24, ..., 3072 form a G.P. with first term a = 3, common ratio r = ~ 2 and last term | 7th term from the end = 2” = sor2(-1) if « 1 3072 : = 3072 = 3072-1). = 82 = 48, Example 2, The 4th term of a GP. is 16 and the 7th term is 128. Find the first term and the common ratio of the GP. (2018) ‘ Solution, Let a be the first term and r be the common ratio of the GP. According to given, a, = 16 and a, = 128 > a=16 ol) and aré = 128 ii) ' Dividing (ii) by (i), we get ars _ 128 - - ee eas =erpa8 >r=2 From (i), ax 23 =16 = 8a=16 => a=2. first term = 2 and common ratio of G.P. is 2. ERA Arithmetic and Geometric ProgressionsFa GP, are 6, 48 and 3072 respectively, Example 3, If the third, sixth and the fast tern sins in the find the first tern and the number of te se the number of terms of the Let ate the frat tore, he common aio and ven GP. According to given, a, = 6, a, = 48 and a, = 3072 6 O} ard = 48 ii) and art! = 3072. 3 From (),ax2=6 > 4a=6 > #= 3 From (ii), 3 x 2"! = 3072 = 2? = 1024 = 20 10 >= 12. ) Example 4, The 4th, 6th and the last terms of a geometric progression are 10, “ 7 60 respectively. If the common ratio is positive, find the first erm common ratio a number of terms of the progression. ) Solution, Let a be the first term, r be the common ratio and 7 be the number of terms of the given GP. According to given, 10, a, = 40 and a, = 640 = aP=10 A) a= 40.) and Dividing (i) by (, we get 640 ii) = ore io =2 but r is positive (given) \| From (i), ax 2 = 10 => 81 = 10 =a | From (ii), 3 2" = 640 = 27 = 640 xt | = ow 512321 =P 9=n=10 = ne Hence, first term = 3, common ratio = 2 and number of terms = 10. | Example 5, The first term of a G.P. with real terms is 1. If the sum of its third and fifth terms is 90, find the common ratio of the G.P. | Solution, Let r(a real number) be the common ratio of the G.P. Given that first term a = 1 and a, + a, 0 = aPsart=9 = P4rt=90 a=0 4-9-0 = (P-9) (2410) =0 « r 9, -10 but r is real so 7% -10 9 a4. Al QUAD GAT (DING ICSE MATHEMATICS -X Shot by RudtarExample 6, If the ple 6, If the sum of the first two terms of a G.P. is 4 a i is 4 ti ele 6 he a te ms of a G.P, is -4 and the fifth term is 4 times Soluti a be the first lution, Let @ be the first term and r be the common ratio of the G.P,, then 4 >atar=-4 >a(l+n=-4 li) a, sart=4xar? = rt=4 > Gpis -4 3 When r = -2, from (i), we get a(1-2)=-4 a= GP. is 4, -8, 16, : Hence, the requi is -4,-8 16 . ane GR is -3,-8,-38, ... oF 4, -8, 16, --- Example 7. The sum of three numbers in GP. is 13 and their product is 1. Find the 2 numbers. Solution, Let three numbers in G.P. be *, a, ar. Their product = : .a.ar = -1 Given) => @®s-l=@=(¢1$ sa=-1 ‘Also sum = 2 +a4ar=2 >a Lert? 13 r 12 5 ‘ 2 derer 1 = (1). 2B 124 r+ WP Lr Dares Br += 9 iS when r= ~3, the terms are 4,- 2 3 Hence, the required numbers are 2 Three numbers whose sum is 21 are in AP. If 2, 2, 14 are added to them the resulting numbers are in G.P. Find the numbers. avd. Example 8. respectively, Solution, Let the three given numbers in AP bea-d,a, Their sum = 32 = 21 => a= 7. ‘Therefore, the numbers are 7 - d, 7, 7 + d. ‘Adding 2, 2, 14 respectively, we get the numbers 9 -d,9, 21 +d. Since, these are in GP, 0% = @-a) QL +d) = @ + 12d - 108-0 $ > +18) @-6)=0 = d= 6 or-18. Using d = 6, we get three numbers as 1, 7 13, and using d = -18, we get three numbers as 25, 7, 11. Hence, the numbers are 1, 7, 13 or 25, 7, -1. The product of first three terms of a GP. is 1000. If 6 is added to its second term ecome in A.P. Find the G-P. Ant 7 add to its third term, then the terms b Solution, Let the first three terms of G-P- be 2, a, ar. a.ar = 1000 (given) ‘nsthmetic and Geometric Progressions> @= = 10. a? = 1000 => a - Gp, are 2, 10, 10F- So, the first three terms of the ‘According to given, 2, 10 + 6, 107 + 7 are in AP 1 16, 10 + 7 are in AP. f 0 = axe 4 aoren => tr 2 4272 2-5r+2=0 = (2) er) =09r=2F When r = 2, the GP. is 5, 10, 20, --- 1, the GP. is 20, 10,5, Hence, the required GP. is 5, 10, 20, - 2x + 2, 3x +3 are first three terms of a ge when r= . oF 20, 10, 5, ++ Example 10. If x, jometric progression, find its 6. fourth term. Solution, As x, 2x + 2, 3x +3 are in GP, (Qx +2) =x x Gx +3) = dx? + Br t= => Pet ds0S +N OtH= -1, 0, 0 which is not possible, as in a G.P. all | Using x = -1, we get first three terms as terms are non-zero. ‘ Using x = -4, we get first three terms as ~4,~6, ~9, which is a G.P, with common ratio “0 3 (Grd term) x common ratio = (-9) (3) --2. Hence, the 4th term | | Example 11. The third term of a GP. is 4. Find the product of its first five terms. 2 | Solution. Let a be the first term and r be the common ratio of given G.P. | Third term = ar? = 4. | Now product of first five terms = a.ar.ar?.ar3.ar* = 95,11 7 . Le | | = (ar?) = (4) = 1024. th | Example 12. If the 4th, 10th and 16th terms of a GP. are x, y and z respectively, prove that “x,y, zare in GP. Solution, Let a be the first term and r be the common ratio of the given GP. ‘Ther = = = mn, 4th term = ar? = x, 10th term = ar? = y and 16th term = arl = 2, | = xy zareinGR 4. (@ Find the next term of the list of numbers (i Find the next term of the list of numbers AITQUADIGAMERA Sirtesel Asie Tall ai(ii) Find the 15th term of the series J3.+ J 4 oR ‘ (io) Find the 10th and nth terms of the list of numbers 5, 25, 125, (©) Find the 6th and the nth terms of the list of numbers 2,3, , (i) Find the 6th term from the end of the list of numbers 3, ~6, 12, -24, ..., 12288 ‘ 2. Which term of the G.P. () 2, 2V2, 4, ... is 1287 12,4)... 2 39 28 Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2 4 Ina GP, the third term is 24 and 6th term is 192. Find the 10th term. 5, Find the number of terms of a G.P. whose first term is 2, common ratio is 2 and the last term is 384. ’ 6, Find the value of x such that iy 2 z . () -F , x, -3 are three consecutive terms of a GP. (ii) x + 9, x-6 and 4 are three consecutive terms of a GP. (ii) x, x +3, x +9 are first three terms of a GP. 7. If the fourth, seventh and tenth terms of a GP. are x, y, z respectively, prove that x, v, zare in GP. . The 5th, 8th and 11th terms of a G.P. are p, q and 5 respectively. Show that q? = ps. 9, Ifa, a +2 and a + 10 are in GP, then find the value(s) of a. “40. Find the geometric progression whose 4th term is 54 and the 7th term is 1458. 11, The sum of first three terms of a GP. is 2 and their product is 1. Find the common ratio and the terms. 42. Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers. 9.2.2 Sum of first » terms of a G.P. Let a be the first term and r the common ratio of the G.P. If S, denotes the sum of first m terms, then S, = atartar+... + art a m= ar+ar?+...+ar™) + ar" Rp Avo CAMERA Sneeloyievicliernill ‘Arithmetic and Geometric ProgressionsRemark Ir=1,then S, = na 2. If Lis the last term of the G.P, then J =a r"~! Thus, 5, = 3 and common ratio r = Example 2. Find the sum of the series 1 - ; ~2 4+. upto 8 terms. | Solution, The given series is a geometric series with first term a = 1 and common ratio 4. Hence, the sum of first 8 terms is ( ‘x’ the number of terms of the GP. (i) sum of n terms. Solution, Let a be the first term of the G.P, then a = 3 (given) and common ratio = r= 2 (@ We know that last term =I = ar! 396 =3x21 = 2 amas sane Hence, the number of terms (given) (i Sum of n terms ie. 6 terms = a =3 x 63 = 189. Example 4, Find the sum of the series Badehe, . The given series is a geometric series with first term a = 2 Soluti & and common ratio 3 and last term 1 = £1, 2 2 ernie IK Slaloisl ane celal Ex 5 seaL 2a" mo 92 2 = (J . The sum of the given series = §, = “7=D ro = Bad s( 221) 4, 2059 _ 2059 Alternatively Seles 9" 128 288 } ; I Using 5, a , the sum of the given series ol 32 | R279 (28s 2059, ae; 3576 288” 2 Note that in this process, there is no need to find the number of terms. 3 3069 Example 5. How many terms of the GP. 3, 34, vate needed to give the sum 5°? = 3 and common ratio r= 2. Let 1 terms of the Solution, The first term of the given GP. given GP. make up the sum ee. Then ~qo2s ~ 1024 = pay = (3)-() snem Hence, the required number of terms = 10. Example 6, Given a GP. with a = 729 and 7th term = 64, determine S,- Solution, Let r be the common ratio of G.P. Then a, = 64 = ar?-1= 64 = 72975 = 4 ‘Shot by RUdranilWhen r= =] 1 @a1s7 + 128) Example 7; Find the number of terms of a geometric progression (ag) if ay = 9 ts = 96 and 5, = 189, Solution, Let r be the common ratio of G.P,, then a, = 0, 71 = 96 = 3.1% = rt = 32 0) S.= ate" 63(r = 1) = 8-1 = Br 62 = (using (#) = Gr-62=1x32 = Blrs@Q>r=2 Substituting this value of r(= 2) in (9), we get amt =32=Bon-1= a n=6 Hence, the number of term: Example 8. The sum of some terms of a GP. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms | Solution. Here a = 5 and r = 2. Let m be the number of terms and / be the last term of the GP. 1x2-5 2-1 Using S, = Et, we get 315 = = 2 =315+5=320 >! = 160. Now | =art-1 = 160 =5 x 2"-* =32=2 > = n-1=5 31=6 Hence, the last term = 160 and the number of terms = 6. Example 9. The sum of first three terms of a GP. is 16 and the sum of the next thr i DE Determine the fist term, common ratio and the sum fo n terms of the GP. Solution, Let a be the first term and r be the common ratio of the given G.P,, then atarsar=16 sal +r+r)=16 ‘ @ . and a + ar +a =128 a3 (14r47)=128 Dividing (ii) by (@, we get Po8 sr=2 Putting this value of r (= 2) in (0, we get a(l+2+4)=16 a= 16. 7 D6e 1. Find the sum of : ( 20 terms of the series 2464 18 +... (ii) 10 terms of series 1+ V3 +34... 24 (iii) 6 terms of the G.P. 1, - (iv) 5 terms and 1 terms of the series 1+ 244 4 2 . 2. Find the sum of the series 81- 27+9-...- 1. Hint anir GP. with a= 81,7 dt=-2 I an Use sum 3 “4. The nth term of a GP. is 128 and the sum of its m terms is 255. If its common ratio is 2, then find its first term. 4. @ How many terms of the GP. 3, 32, 3%, .. are needed to give the sum 120? (ii) How many terms of the G.P. 1, 4, 16, ... must be taken to have their sum equal to 341? 55 + ... will make the sum 7 5, How many terms of the series 2. +& The 2nd and 5th terms of a geometric series are + and 2 respectively. Find the sum of the series upto 8 terms. The first term of a GP. is 27 and 8th term is 1. Find the sum of its first 10 terms. Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728. Ina GP, the first term is 7, the last term is 448, and the sum is 889. Find the common ratio. ~40, Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186. 11, If the first term of ratio. Multiple’ Choice Questions Choose the correct answer from the given four options (1 to 23) : a GP. is 5 and the sum of first three terms is 2, find the common 1, The 10th term of the AP. 5, 8, 11, 14, ... is (a) 32 ( 35 (©) 38 (a) 185 2. The 30th term of the A.P. 10, 7, 4 - 87 (@) 87 () 7 (©) -7 @ - he Uith term of the A.P. -3,-3, 2 : its 48) © -3%8 One (28 & 2 AlLQUAD GAMERAN Shot by Rucranil ‘asthmetic and Geometric Progressionsir —— The 15th term from the last of the AP. 7, 10, 13, ..., 130 is (49 (0) 85 () 88 @ 10 then ayy = Ay 8 ( If the common difference of an A.D. i ws (H) 20 © 25 (d) 30 B 6. Inan AP, if ayy ~ ayy = 92 then the common difference is @s (b) -8 (© -4 (4 7. Inan AP, if d= 4,0 = 7,4, =4, then a is @6 7 (©) 20 (@ 28 8. Inan AP, if a =35,d=0, 1 = 101, then a, will be @o () 35 (©) 1035 (@ 1045 9. Which term of the A.P. 21, 42, 63, 84, ... is 210? (@) 9th (®) 10th (© uth (a) 12th 10. If the last term of the A-P. 5, 3, 1, -1, ... is -41, then the AP. consists of (a) 46 terms (b) 25 terms (c) 24 terms (d) 23 terms M1. Ifk-1,k + 1 and 2k +3 are in AP, then the value of k is @ -2 () 0 2 @4 12. The 21st term of an A-P. whose first two terms are -3 and 4 is @ 17 (b) 137 (©) 143 @ -143 15. If the first term of an A.P. is ~5 and the common difference is 2, then the sum of its first 6 terms is @o 5 @6 @ 15 14, The sum of 25 terms of the AP. -2,-2,-2, ...is 2 50 @o o-2 @-2 @ -50 15, Inan AP, ifa = 1,4, = 20 and §, = 399, then nis @ 19 oa (©) 38 (@ 42 16, In an AP, ifa = ~5,1 = 21 and S = 200, then n is equal to (@ 50 40 © 32 (a) 25 17, The sum of first five multiples of 3 is (@) 45 ©) 55 (0) 65 @ 75 18. The number of two digit numbers which are divisible by 3 is (a) 33 31 (© 30 @ 2% 19, The number of multiples of 4 that lie between 10 and 250 is (a) 62 () 60 (© 59 @ 55 20, The sum of first 10 even whole numbers is (@) 110 90 © 55 @ 45 21, The Ith of the GP. 3-4, 2,—1,... is (a) 64 © - 64 © 28 @ - 128 22, The Sth term from the end of the GP. 2, 6,18, ..., 13122 is (a) 162 () 486 (0) 54 aes 23. If en 1), 3(k + 1) aoe consecutive = of a GP, then the value of k is (c) 1 (@) 4 TATOO GAM Iaidine icse MaTHEMATICS-x Sires elicinilhapter Test 1. Write the first four terms of the A.P. when its first term is -5 and common difference is -3. 2. Verify that each of the following lists of numbers is an AP, and then write its next three terms: 113 ea 4 13 OO Dg 5, Fae 3. The nth term of an A-P. is 6n + 2. Find the common difference. UNDERSTANDING ICSE MATHEMATICS-X4. Show that the list of numbers 9, 12, 15, 18, ... form an A.P. Find its 16th term and the nth term, 5. Find the 6th term from the end of the AP.17, 14, 1, ..., ~40, 6. If the 8th term of an AP. is 31 and the 15th ter find the A.P, The 17th term of an A.P. is 5 more than twice its Bth term. If the Ith term of the AP is 43, then find the nth term. 8. The 19th term of an A.P, the AP, is 16 more than its 1th term, then is equal to three times its 6th term. If its 9th term is 19, find 9. If the 3rd and the 9th terms of an A.P. this A.P. is zero? 10. Which term of the list of numbers 5, 2, 1, -4, ... is 55? ‘1, The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term. are 4 and 8 respectively, then which term of 12. Which term of the list of numbers 20, 19i, 183, Ww, ... is the first negative term? 13, How many three digit numbers are divisible by 92 14, The sum of three numbers in AP. is -3 and the product is 8. Find the numbers. 15. The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles. 16. Find the sum of first 20 terms of an A.P. whose nth term is 15 - 4n. Find the sum: 18 + 152 413+... + (-#93). 18. (@) How many terms of the AP. -6,— (i) Solve the equation 2+5 48+... +x 19. If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is then find the sum of first 20 terms of this A.P. 20. The sum of first 14 terms of an AVP. is 1505 and its first term is 10. Find its 25th term. 21, Find the geometric progression whose 4th term is 54 and 7th term is 1458. The fourth term of a G.P. is the square of its second term and the first term is - 3. Find its 7th term. 23, If the 4th, 10th and 16th terms of a GP. are x, y and z respectively, prove that x, y and zare in GP. 3, . 3069 13, .. are needed to give the sum 27 24, How many terms of the GP. 3, =? ‘Arithmetic and Geometric Progressions — 1 H“oy My S800 (iii) 21978 (in) 816 Exercise 9.4 4 G3 Gi 72 Gi O42 3 (377 (io) 510, 5n 4 n ath i) 6th © Gig i ~3e4 asth (if : a . : P9072... 2072,” 5 10 6. ()lor-l (i) 0orl6 Gii)3 Ae 5 or 2;2,1, 2 0r8,1,2 Mee ma Zorgige gras 12 Le henge Exercise 9.5 | . (93-1 (i) 12103 +1) ii) 133 é 1 @ ) (i) Bey 8, -() | 1640 : | 2 3.1 4. ()4 (5 5.5 F £ | 81 1 7 5(1-J) 8.2 9.2 10, 18, 1. Lor-f Multiple Choice Questions men a ae eee 8. (b) 9. (b) 10. (c) 1. (b) 12. (b) 13. (a) 1.0 15. (0) 16. (@) 17. (a) 18. (c) 19. (b) 20. (6) 21. (0) 22. (@) 23. (6) Chapter Test i 1 1 1, -5,-8,-11,-14 2 (1, 33 (ta) 22s 3.6 4. yg = 54,0, =3n +6 5. -25 6. 3,7, 1,15, 7. An-1 8. 3,5,7,9, 9. 5th 10. 2ist 12. 28th 13. 100 14, -4,-1,2 15. 60°, 80°, 100°, 120° 16. -540 17. ~441 18. (i) 50r20 (ii) x= 29 19. 550 20. 370 24. 10 21. 2,6,18,... 22, -2187