th<\/sup> term in above series.<\/p>\nQuestion 2.
\nFind out the common ratio in the GP 2, 2\\(\\sqrt{2}\\), 4,………..
\nSolution:
\nThe given GP is 2, 2\\(\\sqrt{2}\\), 4, ……….
\nThe common ratio = \\(\\frac{\\mathrm{a}_2}{\\mathrm{a}_1}\\) = \\(\\frac{2 \\sqrt{2}}{2}\\) = \\(\\sqrt{2}\\)<\/p>\n
<\/p>\n
Question 3.
\nThe hand borewell driller charges Rs. 200\/- for the first one meter only and raises drilling charges @ 30\/- for every subsequent meter. Write a progression for the above data. (T.S Mar. 15)
\nSolution:
\nCost of first meter = Rs. 200\/-
\nFor every subsequent meter
\n= Rs. 30\/- raised.
\nSo the progression = 200, 230, 260,………<\/p>\n
Question 4.
\nIn a flower garden there are 23 plants in first row, 21 plants in second row, 19 plants in 3rd row and so on. If there are 10 rows in that garden, then find the to-tal number of plants in the last row with the help of the formula tn<\/sub> = a + (n – 1) d. (T.S. Mar. 16)
\nSolution:
\nNo. of plants in 1st<\/sup> row = 23
\nNo. of plants in 2nd<\/sup> row = 21
\nNo. of plants in 3rd<\/sup> row = 19 and so on.
\nSo the progression is 23, 21, 19, ………..
\nin this A.P a = 23, d = 21 – 23 = – 2
\nn = 10
\ntn<\/sub> = a + (n – 1)
\nt10<\/sub> = 23 + (10 – 1) (-2)
\n= 23 + 9 (-2)
\n= 23 – 18 = 15
\nNumber of plants in the last row = 5.<\/p>\nQuestion 5.
\nFind the sum of first 200 natural numbers. (T.S. Mar. 16)
\nSolution:
\nFormula for the sum of first n natural numbers is \u03a3n<\/sub> = \\(\\frac{\\mathrm{n}(\\mathrm{n}+1)}{2}\\)
\nPut n = 200 in above formula.
\nWe get
\n\u03a3200 = \\(\\frac{200 \\times(200+1)}{2}\\) = \\(\\frac{200 \\times 201}{2}\\)
\n= 20,100<\/p>\nQuestion 6.
\nIf the sides of a triangle are in AP. The perimeter of the triangle is 30 cm. the difference between the longer and shorter side is 4 cm. Then find all sides of the triangle. (T.S. Mar. 16)
\nSolution:
\nLet the 3 sides of given triangle = a – d, a, a + d
\nThen its perimeter
\n= a – d + a + a + d = 30 cm.
\n3a = 30 cm
\n\u21d2 a = \\(\\frac{30}{3}\\) = 10 cm.
\nThe larger side = a + d
\nThe shorter side = a – d
\nThe difference between the above two = (a + d) – (a – d) = 4 cm.
\na + d – a + d = 4 cm.
\n2d = 4; d = \\(\\frac{4}{2}\\) = 2 cm.
\nSo the sides a – d = 10 – 2 = 8 cm
\na = 10 cm
\nand a + d = 10 + 2 = 12 cm.
\nSo 8, 10, 12 cm are the sides of the triangle.<\/p>\n
Question 7.
\nFind the stun of all 3 digit numbers that are divisible by 4.
\nSolution:
\nThe 3 digit numbers are 100, 101, 102,………. 999 among them the number divisible by 4 are 100, 104, 108, … 996 which is an A.P the first term a = 100
\nCommon difference = a2<\/sub> – a1<\/sub>
\n= 104 – 100 = 4
\nLet the number of terms = n
\nThe nth<\/sup> term an<\/sub> = 996
\nan<\/sub> = a + (n – 1) d
\n996 = 100 + (n – 1) 4
\n\\(\\frac{996-100}{4}\\) = n – 1
\n\u21d2 \\(\\frac{896}{4}\\) = n – 1 = 224
\n\u21d2 n = 224 + 1 = 225
\nNow formula for sum of ‘n’ terms in AP is
\nSn<\/sub> = \\(\\frac{\\mathrm{n}}{2}\\)[a + l]
\n= \\(\\frac{225}{2}\\)[100 + 996]
\n= \\(\\frac{225 \\times 1096}{2}\\) = 1, 23, 300<\/p>\nAdditional Questions<\/span><\/p>\nQuestion 1.
\nFor the following A.P’s write the first term and the common difference.<\/p>\n
i) \\(\\frac{1}{4}\\), \\(\\frac{1}{2}\\), \\(\\frac{3}{4}\\), \\(\\frac{5}{4}\\)……….
\nii) 5, 8, 11, 14, 17 …….
\niii) \\(\\frac{1}{3}\\), 1, \\(\\frac{5}{3}\\), \\(\\frac{7}{3}\\),………
\nSolution:
\ni) \\(\\frac{1}{4}\\), \\(\\frac{1}{2}\\), \\(\\frac{3}{4}\\), \\(\\frac{5}{4}\\)……….
\nGiven \\(\\frac{1}{4}\\), \\(\\frac{1}{2}\\), \\(\\frac{3}{4}\\), \\(\\frac{5}{4}\\)……….
\nFirst term = \\(\\frac{1}{4}\\) = a = t1<\/sub>
\nCommon difference = d
\n= t2<\/sub> – t1<\/sub> = \\(\\frac{1}{2}\\) – \\(\\frac{1}{4}\\) = \\(\\frac{1}{4}\\)<\/p>\n<\/p>\n
ii) 5, 8, 11, 14, 17
\nSolution:
\nGiven 5, 8, 11, 14, 17, ………
\nFirst term = 5 = a = t1<\/sub>
\nCommon difference = d
\n= t2<\/sub> – t1<\/sub> = 8 – 5 = 3<\/p>\niii) \\(\\frac{1}{3}\\), 1, \\(\\frac{5}{3}\\), \\(\\frac{7}{3}\\),……..
\nSolution:
\nGiven \\(\\frac{1}{3}\\), 1, \\(\\frac{5}{3}\\), \\(\\frac{7}{3}\\),……
\nFirstterm = \\(\\frac{1}{3}\\) = a = t1<\/sub>
\nCommon difference = d
\n= t2<\/sub> – t1<\/sub>
\n= 1 – \\(\\frac{1}{3}\\) = \\(\\frac{3-1}{3}\\) = \\(\\frac{2}{3}\\)<\/p>\nQuestion 2.
\nWrite the first four terms of the A.P. when the first term ‘a’ and the common’d’ are given as follows.
\ni) a = 6, d = – 2
\nii) a = – 3, d = 4
\niii) a = x + 2y, d = – y
\niv) a = 8, d = 5
\nSolution:
\ni) a = 6, d = – 2
\nGiven a = 6, d = – 2
\nFirst term = t1<\/sub> = a = 6
\nSecond term = t2<\/sub> = a + d = 6 – 2 = 4
\nThird term = t3<\/sub> = a + 2d = 6 + 2(- 2)
\n= 6 – 4 = 2
\nFourth term = t4<\/sub> = a + 3d
\n= 6 + 3 (-2)
\n= 6 – 6 = 0
\n\u2234First four terms are 6, 4, 2, 0<\/p>\nii) a = – 3, d= 4
\nSolution:
\nGiven a = – 3, d = 4
\nFirst term = t1<\/sub> = a = – 3
\nt2<\/sub> = a + d = -3 + 4 = 1
\nt3<\/sub> = a + 2d = -3 + 2 \u00d7 4
\n= -3 + 8 = 5
\nt4<\/sub> = a + 3d
\n= -3 + 3 \u00d7 4
\n= – 3 + 12 = 9
\nFirst four terms :
\n\u2234 First four terms are – 3, 1, 5, 9<\/p>\niii) a = x + 2y, d = – y
\nSolution:
\nGiven a = x + 2y, d = -y
\nFirst term = a = t1<\/sub> = x + 2y
\nt2<\/sub> = a + d
\n= x + 2y – y
\n= x + y
\nt3<\/sub> = a + 2d
\n= (x + 2y) + 2 \u00d7 (- y)
\n= x + 2y – 2y
\n= x
\nt4<\/sub> = a + 3d
\n= (x + 2y) + 3(-y)
\n= x + 2y – 3y
\n= x – y
\n\u2234 First four terms are
\nx + 2y, x + y, x, x – y<\/p>\niv) a = 8, d = 5
\nSolution:
\nGiven a = 8, d = 5
\nFirst term = a = t1<\/sub> = 8
\nt2<\/sub> = a + d
\n= 8 + 5 = 13
\nt3<\/sub> = a + 2d
\n= 8 + 2 \u00d7 5
\n= 8 + 10
\n= 18
\nt4<\/sub> = a + 3d
\n= 8 + 3 \u00d7 5
\n= 8 + 15
\n= 23
\n\u2234 First four terms are 8, 13, 18, 23<\/p>\n<\/p>\n
Question 3.
\nWhich of the following are APs ? If they form an AP, find the common difference and write three more terms.
\ni) 3, 5, 7, 9,……….
\nii) 5, 9, 7, 3, ………..
\niii) 3, \\(\\frac{10}{3}\\), \\(\\frac{11}{3}\\), 4,……….
\niv) 0, – 3, – 6, – 9, -12
\nv) a, 4a, 7a, 10a …………
\nSolution:
\ni) 3, 5, 7, 9,………..
\nGiven 3, 5, 7, 9, …….
\nHere a1<\/sub> = 3, a2<\/sub> = 5, a3<\/sub> = 7
\na2<\/sub> – a1<\/sub> = 5 – 3 = 2
\na3<\/sub> – a2<\/sub> = 7 – 5 = 2
\nSince d = a2<\/sub> – a1<\/sub> = a3<\/sub> – a2<\/sub> = 2 is equal.
\n\u2234 The series form an A.P.
\nNext three terms = 9 + 2 = 11, 11 + 2 = 13, 13 + 2 = 15
\ni.e., 11, 13, 15<\/p>\nii) Given 5, 9, 13, 17
\na1<\/sub> = 5, a2<\/sub> = 9, a3<\/sub> = 13
\na2<\/sub> – a = 9 – 5 = 4
\na3<\/sub> – a2<\/sub> = 13 – 9 = 4
\nSince d = a2<\/sub> – a1<\/sub> = a3<\/sub> – a2<\/sub> = 4 is equal.
\n\u2234The series form an A.P. next three terms
\n17 + 4 = 21
\n21 + 4 = 25
\n25 + 4 = 29
\ni.e., 21, 25, 29<\/p>\niii) 3, \\(\\frac{10}{3}\\), \\(\\frac{11}{3}\\), 4
\nSolution:
\nGiven 3, \\(\\frac{10}{3}\\), \\(\\frac{11}{3}\\), 4,………
\nhere a1<\/sub> = 3, a2<\/sub> = \\(\\frac{10}{3}\\), a3<\/sub> = \\(\\frac{11}{3}\\)
\na2<\/sub> – a1<\/sub> = \\(\\frac{10}{3}\\) – 3 = \\(\\frac{10-9}{3}\\) = \\(\\frac{11}{3}\\)
\na3<\/sub> – a2<\/sub> = \\(\\frac{11}{3}\\) – \\(\\frac{10}{3}\\) = \\(\\frac{11-10}{3}\\) = \\(\\frac{1}{3}\\)
\nSince a2<\/sub> – a1<\/sub> = a3<\/sub> – a2<\/sub> = \\(\\frac{1}{3}\\) = d isequal.
\n\u2234 The series form an AP
\nNext three terms are = 4 + \\(\\frac{1}{3}\\) = \\(\\frac{12-11}{3}\\) = \\(\\frac{13}{3}\\)
\n\\(\\frac{13}{3}\\) + \\(\\frac{1}{3}\\) = \\(\\frac{14}{3}\\), \\(\\frac{14}{3}\\) + \\(\\frac{1}{3}\\) = \\(\\frac{14+1}{3}\\) = \\(\\frac{15}{3}\\) = 5
\ni.e., \\(\\frac{13}{3}\\), \\(\\frac{14}{3}\\), 5<\/p>\niv) 0, -3, -6, -9, -12,……….
\nSolution:
\nGiven 0, – 3, – 6, – 9, – 12,
\nHere a1<\/sub> = 0, a2<\/sub> = – 3, a3<\/sub> = – 6
\na2<\/sub> – a1<\/sub> = -3 – 0 = -3
\na3<\/sub> – a2<\/sub> = -6 – (-3)
\n= -6 + 3 = -3
\nSince a2<\/sub> – a1<\/sub> = a3<\/sub> – a2<\/sub> = – 3 = d is equal.
\n\u2234 The series form an A.P
\nNext three terms are = – 12 – 3
\n= -15
\n= -15 – 3
\n= -18
\n= -18 – 3
\n= -21
\ni.e.,-15,-18,-21<\/p>\nv) a, 4a, 7a, 10a ……….
\nSolution:
\nGiven a, 4a, 7a, 10a, ……….
\nHere a1<\/sub> = a, a2<\/sub> = 4a, a3<\/sub> = 7a
\na2<\/sub> – a1<\/sub> = 4a – a = 3a
\na3<\/sub> – a2<\/sub> = 7a – 4a = 3a
\nSince a2<\/sub> – a1<\/sub> = a3<\/sub> – a2<\/sub> = 3a = d is equal.
\n\u2234 The series form an A.P
\nNext three terms are: 10a + 3a = 13a
\n13a + 3a = 16a, 16a + 3a = 19a
\ni.e., 13a, 16a, 19a.<\/p>\nQuestion 4.
\nFill in the blanks in the following table.
\n
\nSolution:
\n
\n<\/p>\n
<\/p>\n
Question 5.
\nFind the
\ni) 25th<\/sup> term of the A.P.: 8, 11, 14,………
\nii) 10th<\/sup> term of the A.P.
\n– 10, -6, -2,…….
\nSolution:
\ni) Given A.P: 8, 11, 14, ………
\nHere a1<\/sub> = 8, d = a2<\/sub> – a1<\/sub>
\n= 11 – 8 = 3
\nan<\/sub> = a + (n – 1)d
\na25<\/sub> = 8 + (25 – 1) \u00d7 3
\n= 8 + 24 \u00d7 3
\n= 8 + 72
\n= 80<\/p>\nii) Given A.P : – 10, – 6, – 2, ………
\nHere a1<\/sub> = – 10,
\nd = a2<\/sub> – a1<\/sub>
\n= – 6 – (-10)
\n= -6 + 10 = 4
\nan<\/sub> = a + (n – 1) d
\na10<\/sub> = -10 + (10 – 1)4
\n= -10 + (10 – 1)4
\n= -10 + 9 \u00d7 4
\n= – 10 + 36
\n= 26<\/p>\nQuestion 6.
\nWhich term of the A.P 5, 8, 11, 14, ….. is 47?
\nSolution:
\nGiven A.P : 5, 8, 11, 14,……….
\nHere a = 5, d = a2<\/sub> – a1<\/sub> = 8 – 5 = 3
\nLet 47 be the nth term of the given A.P
\n\u2234 an<\/sub> = a + (n – 1) d
\n47 = 5 + (n – 1) 3
\n= 5 + 3n – 3
\n47 = 2 + 3n
\n\u21d2 47 – 2
\n3n = 45
\n\u21d2 n = \\(\\frac{45}{3}\\) = 15
\n\u2234 47 is the 15th term of given A.P<\/p>\nQuestion 7.
\nFind the number of terms of the A.P.
\n7, 12, 17,…… 152.
\nSolution:
\nGiven AP : 7, 12, 17,…….. 152
\nHere a = 7,
\nd = a2<\/sub> – a1<\/sub> = 12 – 7 = 5
\nLet 152 be the nth<\/sup> term of the given A.P
\nThen an<\/sub> = a + (n – 1)d
\n152 = 7 + (n – 1)5
\n152 = 2 + 5n
\n\u21d2 152 – 2 = 5n
\n\u21d2 5n = 150
\n\u21d2 n = \\(\\frac{150}{5}\\)
\n\u2234 30 terms are there in the given AP<\/p>\nQuestion 8.
\nFind the 21st<\/sup> term of an A.P Whose 11th<\/sup> term is 92 and 16th<\/sup> term is 122.
\nSolution:
\nGiven an AP whose
\n
\nd = \\(\\frac{30}{5}\\) = 6
\nSubstituting d = 6 in equation (1)
\na + 10 \u00d7 6 = 92
\na + 60 = 92
\n\u21d2 a = 92 – 60 = 32
\nNow the 21st<\/sup> term = a + 20d
\n= 32 + 20 \u00d7 6
\n= 32 + 120 = 152<\/p>\nQuestion 9.
\nFind the sum of the following APs.
\ni) 4, 9, 14, …… to 14 terms
\nii) – 32, – 28, – 24, …… to 12 terms.
\nSolution:
\ni) Given 4, 9, 14, …….. to 14 terms
\nHere a = 4, d = a2<\/sub> – a1<\/sub> = 9 – 4 = 5
\nn = 14
\nSn<\/sub> = \\(\\frac{n}{2}\\)[2a + (n – 1)d]
\nS14<\/sub> = \\(\\frac{14}{2}\\)[2 \u00d7 4 + (14 – 1) \u00d7 5]
\n= 7 [8 + 13 \u00d7 5]
\n= 7 [8 + 65]
\n= 7 \u00d7 73
\n= 511<\/p>\n<\/p>\n
ii) – 32, – 28, – 24, ……. to 12 terms.
\nSolution:
\nGiven -32, – 28, – 24, …… to 12 terms
\nHere a = – 32, d = a2<\/sub> – a1<\/sub>
\n= -28 – (- 32)
\n= – 28 + 32 = 4
\nand n = 12
\nSn<\/sub> = \\(\\frac{n}{2}\\)[2a + (n – 1)d]
\nS12<\/sub> = \\(\\frac{12}{2}\\) [2 \u00d7 (-32) + (12 – 1) \u00d7 4]
\n= 6 [-64 + 11 \u00d7 4]
\n= 6[-20]
\n= -120<\/p>\nQuestion 10.
\nIn an A.P. given a = 5, d = 6, an<\/sub> = 89, find n and Sn<\/sub>.
\nSolution:
\nGiven a = 5, d = 6, an<\/sub> = 89
\n\u21d2 a + (n – 1) d = 89
\n\u21d2 5 + (n – 1) 6 = 89
\n\u21d2 5 + 6n – 6 = 89
\n\u21d2 6n – 1 = 89
\n\u21d2 6n = 89 + 1 = 90
\nSn<\/sub> = \\(\\frac{n}{2}\\)[a + l]
\n= \\(\\frac{15}{2}\\)[5 + 89]
\n= 15 \u00d7 47 = 705
\n\u2234 S15<\/sub> = 705<\/p>\nQuestion 11.
\nIn an A.P. given a = 6, a13<\/sub> = 12 = (l) find d and S13<\/sub>.
\nSolution:
\nGiven a = 6, a13<\/sub> = 12
\na13<\/sub> = a + 12d = 12
\n6 + 12 \u00d7 d = 12
\n\u21d2 12d = 12 – 6 = 6
\n\u21d2 d = \\(\\frac{6}{12}\\) = \\(\\frac{1}{2}\\)
\nNow Sn<\/sub> = \\(\\frac{n}{2}\\)[a + l]
\nS13<\/sub> = \\(\\frac{13}{2}\\)[6+12]
\nS13<\/sub> = \\(\\frac{13}{2}\\)[18] = 13 \u00d7 9 = 117<\/p>\nQuestion 12.
\nIn an A.P. given a14<\/sub> = 57, d = 4. Find ‘a’ and S10<\/sub>.
\nSolution:
\nGiven a14<\/sub> = a + 13d = 57 (= l) and d = 4
\na + 13 \u00d7 4 = 57
\na + 52 = 57
\n\u21d2 a = 57 – 52 = 5
\nNow Sn<\/sub> = \\(\\frac{n}{2}\\)[a + l]
\nS10<\/sub> = \\(\\frac{10}{2}\\)[5+5]
\n= 5 [62]
\nS10<\/sub> = 310<\/p>\nQuestion 13.
\nIn an A.P. an<\/sub> = 10, d = 3, Sn<\/sub> = 15 find n and ‘a’.
\nSolution:
\nGiven an<\/sub> = a + (n – 1) d = 10,
\nd = 3, Sn<\/sub> = 15
\n\u21d2 a + (n – 1) 3 = 10
\n\u21d2 a + 3n – 3 = 10
\n\u21d2 a + 3n = 13
\n\u21d2 a = 13 – 3 n
\nSn<\/sub> = \\(\\frac{n}{2}\\)[a + an<\/sub>]
\n= \\(\\frac{n}{2}\\)[13 – 3n + 10]
\n15 \u00d7 2 = n[23 – 3n]
\n30 = 23n – 3n2<\/sup>
\n\u21d2 3n2<\/sup> – 23n + 30 = 0
\n\u21d2 3n2<\/sup> – 18n – 5n + 30 = 0
\n\u21d2 3n(n – 6) – 5(n – 6) = 0
\n\u21d2 (n – 6) (3n – 5) = 0
\n\u21d2 n – 6 = 0 or 3n – 5 = 0
\n\u2234 n = 6; a = \\(\\frac{15}{3}\\) (n cannot be fraction)
\na = 13 – 3n
\n= 13 – 3 \u00d7 6
\na = 13 – 18 = – 5
\n\u2234 a = – 5, n = 6<\/p>\nQuestion 14.
\nIf the sum of first 7 terms of an A.P is 77 and that of 17 terms is 442, find the sum of first “n” terms.
\nSolution:
\nGiven A.P. such that
\nS7<\/sub> = 77
\nS17<\/sub> = 442
\nWe know that Sn<\/sub> = \\(\\frac{n}{2}\\)[2a + (n – 1) d]
\nS7<\/sub> = 77
\n\\(\\frac{7}{2}\\)[2a + (7 – 1)d] = 77
\n\u21d2 2a + 6d = \\(\\frac{77 \\times 2}{7}\\)
\n\u21d2 2a + 6d = 22 (Dividing by 2)
\n\u21d2 a + 3d = 11 —– (1)
\nS17<\/sub> = 442
\n\\(\\frac{17}{2}\\)[2a + (17 – 1)d] = 442
\n\u21d2 2a + 16d = 52 (Dividing by 2)
\n\u21d2 a + 8d = 26 —— (2)
\n
\nSubstitute d = 3 in equation (1)
\na + 3 \u00d7 3 = 11
\n\u21d2 a + 9 = 11
\n\u21d2 a = 11 – 9 = 2
\n\u2234 a = 2, d = 3
\nNow Sn<\/sub> = \\(\\frac{n}{2}\\)[2a + (n – 1) d]
\n= \\(\\frac{n}{2}\\)[2 \u00d7 2 + (n – 1)3]
\n= \\(\\frac{n}{2}\\)[4 + 3n – 3]
\nSn<\/sub> = \\(\\frac{n}{2}\\)[3n + 1]
\n\u2234 Sum of first n terms = Sn = \\(\\frac{n}{2}\\)(3n + 1)<\/p>\n<\/p>\n
Question 15.
\nWrite the terms of the G.P. When the first term ‘a’ and the common ratio ‘r’ are given.
\ni) a = 5, r = 2
\nii) a = \\(\\sqrt{3}\\), r = \\(\\frac{1}{2}\\)
\niii) a = 16, r = –\\(\\frac{1}{2}\\)
\nSolution:
\ni) a = 5, r = 2
\nThe terms of G.P are a, ar, ar2<\/sup>, ar3<\/sup>, ……….
\ni.e., 5, 5 \u00d7 2, 5 \u00d7 22<\/sup>, 5 \u00d7 23<\/sup>,………..
\n\u21d2 5, 10, 20, 40, ……..<\/p>\nii) a = \\(\\sqrt{3}\\), r = \\(\\frac{1}{2}\\)
\nThe terms of G.P are a, ar, ar2<\/sup>, ar3<\/sup>, ……….
\ni.e., \\(\\sqrt{3}\\), \\(\\sqrt{3}\\) \u00d7 \\(\\frac{1}{2}\\), \\(\\sqrt{3}\\) \u00d7 \\(\\frac{1}{2^2}\\), \\(\\sqrt{3}\\) \u00d7 \\(\\frac{1}{2^3}\\)<\/p>\niii) a = 16, r = –\\(\\frac{1}{2}\\)
\nThe terms of G.P are a, ar, ar2<\/sup>, ar3<\/sup>, ………
\ni.e., 16, 16 \u00d7 \\(\\left(\\frac{-1}{2}\\right)\\), 16 \u00d7 \\(\\left(\\frac{-1}{2}\\right)^2\\), 16 \u00d7 \\(\\left(\\frac{-1}{2}\\right)^3\\) ………
\n\u21d2 16, -18, 4, -2,……..<\/p>\nQuestion 16.
\nWhich of the following are GP ? If them are G.P., write 3 more terms.<\/p>\n
i) 3, 15, 75,………
\nii) \\(\\frac{1}{2}\\), –\\(\\frac{1}{6}\\), \\(\\frac{1}{18}\\),………
\niii) a = \\(\\frac{1}{2}\\), r = –\\(\\frac{1}{3}\\)
\niv) – 5 – 10, – 20,………
\nv) – 0.3, – 0.03, – 0.003,………..
\nSolution:
\ni) Given 3, 15, 75, ………
\nWhen a1<\/sub> = 3, a2<\/sub> = 15, a3<\/sub> = 75
\n\\(\\frac{\\mathrm{a}_2}{\\mathrm{a}_1}\\) = \\(\\frac{15}{3}\\) = 5, \\(\\frac{\\mathrm{a}_3}{\\mathrm{a}_2}\\) = \\(\\frac{75}{15}\\) = 5
\n\u2234 r = \\(\\frac{\\mathrm{a}_2}{\\mathrm{a}_1}\\) = \\(\\frac{\\mathrm{a}_3}{\\mathrm{a}_2}\\) = 5
\nHence 3, 15, 75, ……. is a G.P
\nWhen a = 3, r = 5
\na4<\/sub> = a.r3<\/sup> = 3 \u00d7 (5)3<\/sup> = 3 \u00d7 125 = 375
\na5<\/sub> = a.r4<\/sup> = 3 \u00d7 (5)4<\/sup> = 3 \u00d7 625 = 1,875
\na6<\/sub> = a.r5<\/sup> = 3 \u00d7 (5)5<\/sup> = 3 \u00d7 3125 = 9,375<\/p>\nii) \\(\\frac{1}{2}\\), –\\(\\frac{1}{6}\\), \\(\\frac{1}{18}\\),………
\nSolution:
\nGiven a1<\/sub> = \\(\\frac{1}{2}\\), a2<\/sub> = –\\(\\frac{1}{6}\\), a3<\/sub> = \\(\\frac{1}{18}\\)
\n\\(\\frac{\\mathrm{a}_2}{\\mathrm{a}_1}\\) = \\(\\frac{-1 \/ 6}{1 \/ 2}\\) = \\(\\frac{-1}{6}\\) \u00d7 \\(\\frac{2}{1}\\) = \\(\\frac{-1}{3}\\)
\n\\(\\frac{\\mathrm{a}_3}{\\mathrm{a}_2}\\) = \\(\\frac{1 \/ 18}{-1 \/ 6}\\) = \\(\\frac{1}{18}\\) \u00d7 \\(\\frac{-6}{1}\\) = \\(\\frac{-1}{3}\\)
\nHere, \\(\\frac{a_2}{a_1}\\) = \\(\\frac{a_3}{a_2}\\) = r = \\(\\frac{-1}{3}\\)
\nGiven terms are in G.P.<\/p>\n<\/p>\n
iii) a = \\(\\frac{1}{2}\\), r = –\\(\\frac{1}{3}\\)
\nSolution:
\n<\/p>\n
iv) -5, -10, -20,………
\nSolution:
\nGiven a1<\/sub> = -5, a2<\/sub> = -10, a3<\/sub> = -20
\n\\(\\frac{a_2}{a_1}\\) = \\(\\frac{-10}{-5}\\) = 2, \\(\\frac{a_3}{a_2}\\) = \\(\\frac{-20}{-10}\\) = -2
\nSince \\(\\frac{a_2}{a_1}\\) = \\(\\frac{a_3}{a_2}\\) = 2 = r
\n\u2234 Given is in G.P
\na4<\/sub> = ar3<\/sup> = -5 \u00d7 (2)3<\/sup> = -5 \u00d7 8 = -40
\na5<\/sub> = ar4<\/sup> = -5 \u00d7 (2)4<\/sup> = -5 \u00d7 16 = -80
\na6<\/sub> = ar5<\/sup> = -5 \u00d7 (2)5<\/sup> = -5 \u00d7 32 = -160<\/p>\nv) 0.3, 0.03, 0.003,……..
\nSolution:
\nGiven a1<\/sub> = 0.3, a2<\/sub> = 0.03, a3<\/sub> = 0.003
\nSince \\(\\frac{a_2}{a_1}\\) = \\(\\frac{0.03}{0.3}\\) = 0.1, \\(\\frac{a_3}{a_2}\\) = \\(\\frac{0.003}{0.03}\\) = 0.1
\n\\(\\frac{a_2}{a_1}\\) = \\(\\frac{a_3}{a_2}\\) = 0.1 = r
\n\u2234 Given is in GP
\na4<\/sub> = ar3<\/sup> = 0.3 \u00d7 (0.1)3<\/sup> = 0.0003
\na5<\/sub> = ar4<\/sup> = 0.3 \u00d7 (0.1)4<\/sup> = 0.00003
\na6<\/sub> = ar5<\/sup> = 0.3 \u00d7 (0.1)5<\/sup> = 0.000003<\/p>\n<\/p>\n
Question 17.
\nFor each geometric progression find the common ratio \u2018r\u2019 and then find an<\/sub>.<\/p>\ni) 2, \\(\\frac{2}{3}\\), \\(\\frac{2}{9}\\), \\(\\frac{2}{27}\\),……..
\nii) -3, -6, -12, -24,……..
\nSolution:
\ni) Given G.P. : 2, \\(\\frac{2}{3}\\), \\(\\frac{2}{9}\\), \\(\\frac{2}{27}\\),………
\nHere a = 2, r = \\(\\frac{a_2}{a_1}\\) = \\(\\frac{2}{\\frac{3}{2}}\\) = \\(\\frac{1}{3}\\)
\nan<\/sub> = arn-1<\/sup> = 2.\\(\\left(\\frac{1}{3}\\right)^{\\mathrm{n}-1}\\)<\/p>\n