{"id":215,"date":"2021-07-24T16:00:41","date_gmt":"2021-07-24T10:30:41","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=215"},"modified":"2021-07-24T16:10:27","modified_gmt":"2021-07-24T10:40:27","slug":"kseeb-sslc-class-10-maths-solutions-chapter-1-ex-1-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-1-ex-1-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3<\/strong> are part of KSEEB SSLC Class 10 <\/a>Maths<\/a> Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3<\/h2>\n

Arithmetic Progression Class 10 Exercise 1.3 Solutions KSEEB Question 1.<\/strong>
\nFInd the sum of the following APs:
\ni) 2, 7, 12, ……… to 10 terms
\nii) -37, -33, -29,………. to 12 terms.
\niii) 0.8, 1.7, 2.8 to 100 terms.
\niv) \\(\\frac{1}{15}, \\frac{1}{12}, \\frac{1}{10} \\ldots, \\text { to } 11\\) terms
\nSolution:
\ni) Sum of 2, 7, 12, ………. 10?
\na = 2, d = 7 – 2 = 5, n = 10
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(s_{10}=\\frac{10}{2}[2 \\times 2+(10-1) 5]\\)
\n= 5[4 + 9 \u00d7 5]
\n= 5[4 + 45]
\n= 5 \u00d7 49
\n\u2234 S10<\/sub> = 245<\/p>\n

ii) -37, -33, -29, to 12……. terms.
\nSolution:
\na = -37, d = -33 – (-37) = -33 + 37
\nn = 12, d = 4
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{12}=\\frac{12}{2}[2 \\times -37+(12-1) 4]\\)
\n= 6[-74 + 11 \u00d7 4]
\n= 6[-74 + 44]
\n= 6 \u00d7 -30
\n\u2234 S12<\/sub> = -180.<\/p>\n

iii) 0.6, 1.7, 2.8,……… , to 100 terms.
\na = 0.6, d = 1.7 – 0.6 = 1.1
\nn = 100, S100<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{100}=\\frac{100}{2}[2 \\times 0.6+(100-1)(1.1)]\\)
\n= 50[1.2 + 99(1.1)]
\n= 50[1.2 + 108.9]
\n= 50 \u00d7 110.1
\n\u2234 S100<\/sub> = 5505<\/p>\n

iv) \\(\\frac{1}{15}, \\frac{1}{12}, \\frac{1}{10} \\ldots, \\text { to } 11\\) terms
\nSolution:
\n\"KSEEB<\/p>\n

Arithmetic Progression Exercise 1.3 KSEEB Soltion Question 2.<\/strong>
\nFind the sums given below:
\ni) \\(7+10 \\frac{1}{2}+14+\\ldots \\ldots+84\\)
\nii) 34 + 32 + 30 + ……….+ 10
\niii) -5 + (-8) + (-11) + ……….. +(-230)
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

ii) 34 + 32 + 30 + ………. + 10
\nSolution:
\na = 34, d = 32 – 34 =-2, an<\/sub> = 10,
\nSn<\/sub> =?
\na + (n – 1)d = an<\/sub>
\n34 + (n – 1)(-2) = 10
\n34 – 2n + 2 = 10
\n-2n + 36 = 10
\n-2n = 10 – 36
\n-2n = -26
\n2n = 26
\n\\(n=\\frac{26}{2}\\)
\n\u2234 n = 13
\n\\(S_{n}=\\frac{n}{2}[a+l]\\)
\n\\(\\mathrm{S}_{23}=\\frac{13}{2}[34+10]\\)
\n\\(=\\frac{13}{2} \\times 44\\)
\n= 13 \u00d7 22
\n\u2234 S23<\/sub> = 286.<\/p>\n

iii) -5 + (-8) + (-11) +……… + (-230)
\nSolution:
\na = -5, d = -8 – (-5) = -8 + 5
\na, = -230.
\nSn<\/sub> =? d = -3
\na+(n – 1)d = an<\/sub>
\n-5 + (n – 1)(-3) = -230
\n-5 – 3n + 3 = -230
\n-3n – 2 = -230
\n-3n = -230 +2
\n-3n = -228
\n3n = 228
\n\u2234 n = 228\/3
\n\u2234 n = 76.
\n\\(\\mathrm{S}_{\\mathrm{n}}=\\frac{\\mathrm{n}}{2}\\left[\\mathrm{a}+\\mathrm{a}_{\\mathrm{n}}\\right]\\)
\n\\(S_{ 76 }=\\frac { 76 }{ 2 } \\left[ -5-230 \\right] \u00a0\\)
\n\\(=\\frac{76}{2} \\times -235\\)
\n= 38 \u00d7 -235
\n\u2234 S76<\/sub>= -8930.<\/p>\n

KSEEB Solutions For Class 10 Maths Arithmetic Progression Exercise 1.3 Question 3.<\/strong>
\nIn an AP:
\ni) given a = 5. d = 3. an<\/sub> = 50, find \u2018n\u2019 and Sn<\/sub>.
\nii) given a = 7, a13<\/sub> = 35, find d\u2019 and S13<\/sub>.
\niii) given a12<\/sub>= 37. d = 3. find \u2018a\u2019 and S12<\/sub>.
\niv) given a3<\/sub> = 15, S10<\/sub> = 125. find \u2018d\u2019 and a10<\/sub>.
\nv) given d = 5. S9<\/sub> = 72, fInd \u2018a\u2019 and a9<\/sub>.
\nvi) given a = 2. d = 8, Sn<\/sub> = 90, fInd \u2018n and a11<\/sub>.
\nvii) given a = 8, an<\/sub> = 62. Sn<\/sub> = 210, find \u2018n\u2019 and \u2018d\u2019.
\nviii) given an<\/sub> = 4, d = 2, Sn<\/sub> = -14. find \u2018n\u2019 and a.
\nix) given a = 3, n = 8, S = 192, find \u2018d\u2019.
\nx) given L = 28, S = 144 and there are total 9 terms. Find \u2018a\u2019.
\nSolution:
\ni) a = 5, d = 3, an<\/sub> = 50. n=?, Sn<\/sub> =?
\na + (n – 1)d = an<\/sub>
\n5 + (n – 1) 3 = 50
\n5 + 3n – 3 = 50
\n3n + 2 = 50
\n3n = +50 – 2
\n3n = 48
\n\\(n=\\frac{48}{3}\\)
\n\u2234 n = 16
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(S_{16}=\\frac{16}{2}[5+50]\\)
\n\\(=\\frac{16}{2}[55]\\)
\n= 8 \u00d7 55
\n\u2234 S16<\/sub>= 440.<\/p>\n

ii) a = 7, a13<\/sub> = 35. d =?. S13<\/sub> =?
\nSolution:
\na+(n – 1) d = an<\/sub>
\n7 + (13 – 1) d = 35
\n7 + 12d = 35
\n12d = 35 – 7
\n12d = 28
\n\\(d=\\frac{28}{12}\\)
\n\\(d=\\frac{7}{3}\\)
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(S_{n}=\\frac{7}{3} \\times \\frac{1}{2}[7+35]\\)
\n\\(\\mathrm{s}_{\\mathrm{n}}=\\frac{7}{6} \\mathrm{I} 42 \\mathrm{l}\\)
\n= 7 \u00d7 7
\n\u2234 Sn<\/sub> = 49<\/p>\n

iii) a12<\/sub>= 37, d = 3, a =? S12<\/sub> =?
\na + (n – 1)d = an<\/sub>
\na + (12 -1) 3 = 37
\na + 11 \u00d7 3 = 37
\na + 33 = 37
\n\u2234 a = 37 – 33
\n\u2234 a = 4
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(\\mathrm{S}_{12}=\\frac{12}{2}[4+37]\\)
\n= 6[4 + 37]
\n\u2234 S2<\/sub> = 246<\/p>\n

iv) a3<\/sub> = 15, S10<\/sub> = 125. d =?, a10<\/sub>=?
\nSolution:
\na3<\/sub> = a + 2d = 15
\n\u2234 a= 15 – 2d
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(s_{10}=\\frac{10}{2}[2(15-2 d) \\div(10-1) d]=125\\)
\n= 5[30 – 4d + 9d] = 125
\n= 5[30 + 5d] = 125
\n150 + 25d = 125
\n25d = 125 + 150
\n25d = -25
\n\\(\\mathrm{d}=\\frac{-25}{25}\\)
\n\u2234 d = -1.
\nSubstittuting the value of \u2018d\u2019
\na = 15 – 2d
\n= 15 – 2(-1)
\n= 15 + 2
\n\u2234 a = 17.
\n\u2234 a10<\/sub> = a + 9d
\n= 17 + 9(-1)
\n= 17 – 9
\n\u2234 a10<\/sub> = 8<\/p>\n

v) d = 5, S9<\/sub> = 72. a =? a9<\/sub> =?
\nSolution:
\n\\(\\mathrm{s}_{\\mathrm{n}}=\\frac{\\mathrm{n}}{2}[2 \\mathrm{a}+(\\mathrm{n}-1) \\mathrm{d}]\\)
\n\\(s_{9}=\\frac{9}{2}[2 \\times a+(9-1) 5]=72\\)
\n\\(\\frac{9}{2}[2 a+8 \\times 5]=72\\)
\n\\(\\frac{9}{2}[2 a+40]=72\\)
\n\\(2 a+40=72 \\times \\frac{2}{9}\\)
\n2a + 40 = 16
\n2a = 16 – 40
\n2a = -24
\n\\(a=\\frac{-24}{2}\\)
\n\u2234 a = -12
\na9<\/sub> = a + (n – 1) d
\n= -12 + (9 – 1) (5)
\n= -12 + 8 \u00d7 5
\n= -12 + 40
\n\u2234 a9<\/sub> = 28<\/p>\n

vi) a = 2, d = 8 Sn<\/sub> = 90, n =?, a3<\/sub> =?
\nSolution:
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(=\\frac{n}{2}[2 \\times 2+(n-1) 8]=90\\)
\n\\(\\frac{n}{2}[4+8 n-8]=90\\)
\nn (8n – 4) = 90 \u00d7 2\/1
\n8n2<\/sup> – 4n = 180
\n8n2<\/sup> – 4n = 1800
\n2n2<\/sup> – n – 45 = 0
\n\"KSEEB
\n2n2<\/sup>– 10n + 9n – 45 = 0
\n2n(n – 5) + 9(n – 5) = 0
\n(n – 5)(2n + 9) = 0
\nIf n – 5 = 0 then n = 5
\n\u2234 an<\/sub> = a + (n – 1) d
\na5<\/sub>= 2 + (5 – 1) 8
\n= 2 + 4 \u00d7 8
\n= 2 + 32
\n\u2234 a5<\/sub> = 34<\/p>\n

vii) a = 8, an<\/sub> = 62, Sn<\/sub> = 210, n =? d =?
\nSolution:
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]=210\\)
\n\\(\\frac{n}{2}[8+62]=210\\)
\n\\(\\frac{n}{2} \\times 70=21\\)
\n\\(n=210 \\times \\frac{2}{70}\\)
\n\u2234 n = 6
\nan<\/sub> = 62
\na6<\/sub>= a + 5d = 62
\n8 + 5d = 62
\n5d = 62 – 8
\n5d = 54
\n\\(d=\\frac{54}{5}\\)<\/p>\n

viii) a = 4, d = 2, Sn<\/sub> = -14, n =?, a =?
\nSolution:
\nan<\/sub> = a + (n – 1) d = 4
\na + (n – 1) 2 = 4
\na + 2n – 2 = 4
\na + 2n = 4 + 2
\na = 2n = 6 …………. (i)
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]=210\\)
\n\\(\\frac{n}{2}[-2 n+6+4]=-14\\)
\n\\(\\frac{n}{2}[-2 n+10]=-14\\)
\n-2n2<\/sup> + 10n = -28
\n-2n2<\/sup> + 10n + 28 = 0
\n2n2<\/sup> – 10n – 28 = 0 – 14
\n2n<\/span>2<\/sup> – 5n – 14 = 0
\n\"KSEEB
\nn2<\/sup> – 7n + 2n – 14 = 0
\nn(n – 7) + 2 (n – 7) = 0
\n(n – 7) (n + 2) = 0
\nIf n – 7 = 0 then. n = 7
\na = -2n + 6
\n= -2 \u00d7 7 + 6
\n= -14 + 6
\n\u2234 a = -8
\n\u2234 n = 7, a = -8.<\/p>\n

ix) a = 3, n = 8, S = 192, d =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{8}=\\frac{8}{2}[2 \\times 3+(8-1) d]=192\\)
\n\\(\\frac { 8 }{ 2 } [6+7\\times d]=192\\)
\n4 (6 + 7d) = 192
\n\\(6+7 d=\\frac{192}{4}\\)
\n6 + 7d = 48
\n7d = 48 – 6
\n7d = 42
\n\\(d=\\frac{42}{7}\\)
\n\u2234 d = 6<\/p>\n

x) l = an<\/sub> = 28, S = 144, n = 9, a =?
\nSolution:
\n\\(s_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(s_{9}=\\frac{9}{2}[a+28]=144\\)
\n\\(\\frac{9}{2}[a+28]=144\\)
\n\\(a+28=144 \\times \\frac{2}{9}\\)
\na + 28 = 32
\na = 32 – 28
\n\u2234 a = 4.<\/p>\n

Exercise 1.3 Class 10 Maths Arithmetic Progression KSEEB Solutions Question 4.<\/strong>
\nHow many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
\nSolution:
\n9 + 17 + 25 + ……. + an<\/sub> = 636
\na = 9, d = 17 – 9 = 8, Sn<\/sub>= 636, n =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{n}=\\frac{n}{2}[2 \\times 9+(n-1) 8]=636\\)
\n\\(\\frac{n}{2}[18+8 n-8]=636\\)
\n\\(\\frac{n}{2}[8 n+10]=636\\)
\nn(8n + 10) = 636 \u00d7 2
\n8n2<\/sup> + 10n = 1272
\n8n2<\/sup>+ 10n – 1272 = 0
\n4n2<\/sup> + 5n – 636 = o
\n4n2<\/sup>+ 5n – 48n – 636 = 0
\n\"KSEEB
\nn(4n + 5) – 12(4n + 5) = 0
\n(4n + 5) (n – 12) = 0
\n1f n – 12 = 0 then, n = 12
\n\u2234 n = 12.<\/p>\n

KSEEB Solutions For Class 10 Maths Arithmetic Progression Question 5.<\/strong>
\nThe first term of an AP is 5. the last term is 45 and the sum is 400. Find the number of terms and the common difference.
\nSolution:
\na = 5, an<\/sub> = 45, Sn<\/sub>= 400, n =?, d =?
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(=\\frac{n}{2}[5+45]=400\\)
\n\\(\\frac{n}{2} \\times 50=400\\)
\nn \u00d7 50 = 800
\n\\(n=800 \\times \\frac{1}{50}\\)
\n\u2234 n = 16
\nan<\/sub> = a + (n – 1) d
\na16<\/sub>= 5 + (16 – 1)d = 45
\n5 + 15d = 45
\n15d = 45 – 5
\n15d = 40
\n\\(d=\\frac{40}{15}=\\frac{8}{3}\\)
\n\\(\\mathrm{n}=16, \\mathrm{d}=\\frac{8}{3}\\)<\/p>\n

Arithmetic Progression Class 10 Exercise 1.3 KSEEB Solutions Question 6.<\/strong>
\nThe first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
\nSolution:
\na = 17, an<\/sub>= 350, d = 9, n =?, Sn<\/sub> =?
\nan<\/sub> = a + (n – 1) d
\n17 + (n – 1)9 = 350
\n17 + 9n – 9 = 350
\n9n + 8 = 350
\n9n = 350 – 8
\n9n = 342
\n\\(\\quad n=\\frac{342}{9}\\)
\n\u2234 n = 38
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(S_{38}=\\frac{38}{2}[17+350]\\)
\n= 19 \u00d7 367
\n\u2234 S38<\/sub> = 6973
\n\u2234 n = 38, S38<\/sub> = 6973<\/p>\n

10th Maths Arithmetic Progression Exercise 1.3 KSEEB Solutions Question 7.<\/strong>
\nFind the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
\nSolution:
\nd = 7, a22<\/sub> = 149, S22<\/sub> =?
\na = a + (n – 1)d
\na + (22 – 1)7 = 149
\na + 21 \u00d7 7 = 149
\na + 147 = 149
\n=149 – 147
\n\u2234 a = 2
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(S_{22}=\\frac{22}{2}[2+149]\\)
\n= 11[151]
\n\u2234 S22<\/sub> = 1661<\/p>\n

10th Standard Maths Arithmetic Progression Exercise 1.3 KSEEB Solutions Question 8.<\/strong>
\nFind the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
\nSolution:
\nIf a2<\/sub> = 14, a3<\/sub> = 18, then S11<\/sub> =?
\na2<\/sub> = 14
\na + d = 14 …………. (i)
\na3<\/sub> = 18
\na + 2d = 18 …………… (ii)
\nfrom equation (1) – equatIon (11),
\n\"KSEEB
\nd =4.
\nSubstituting the value of \u2018d\u2019 In eqn. (i)
\na + d = 14
\na + 4 = 14
\na = 14 – 4
\n\u2234 a = 10
\na = 10, d = 4
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{51}=\\frac{51}{2}[2 \\times 10+(51-1) 4]\\)
\n\\(=\\frac{51}{2}[20+50 \\times 4]\\)
\n\\(=\\frac{51}{2}[20+200]\\)
\n\\(=\\frac{51}{2} \\times 220\\)
\n= 51 \u00d7 110
\n\u2234 S51<\/sub> = 5610<\/p>\n

10th Arithmetic Progression Exercise 1.3 KSEEB Solutions Question 9.<\/strong>
\nIf the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first \u2018n terms.
\nSolution:
\nS7<\/sub> = 49, S17<\/sub> = 289, S11<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{7}=\\frac{7}{2}[2 a+(7-1) d]\\)
\n\\(=\\frac{7}{2}[2 a+6 d]=49\\)
\n[latex2+6 d=49 \\times \\frac{2}{7}[\/latex]
\n2a + 6d = 14 …………… (i)
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{17}=\\frac{17}{2}[2 a+(n-1) d]\\)
\n\\(=\\frac{17}{2}[2 \\mathrm{a}+16 \\mathrm{d}]\\)
\n\\(\\frac{17}{2}[2 a+16 d]=289\\)
\n\\(2 a+16 d=289 \\times \\frac{2}{17}\\)
\n\u2234 2a + 16d = 34 …………. (ii)
\n\"KSEEB
\n10d = 20
\n\u2234 d=2
\nSubstituting the value of d in equation (i), we have
\n2a + 6d = 14
\n2a + 6 \u00d7 2 = 14
\n2a + 12 = 14
\n2a = 14 – 12
\n2a = 2
\n\\(a=\\frac{2}{2}=1\\)
\na = 1, d = 2, Sn<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{n}=\\frac{n}{2}[2 \\times 1+(n-1) 2]\\)
\n\\(=\\frac{\\mathrm{n}}{2}[2+2 \\mathrm{n}-2]\\)
\n\\(=\\quad \\frac{\\mathrm{n}}{2}[2 \\mathrm{n}]\\)
\n\u2234 Sn<\/sub> = n2<\/sup><\/p>\n

Maths Arithmetic Progression Exercise 1.3 KSEEB Solutions Question 10.<\/strong>
\nShow that a1<\/sub>, a2<\/sub>, a3<\/sub>, … a11<\/sub>. … form an AP where a11 is defIned as below:
\ni) a = 3 + 4n
\nii) a = 9 – 5n
\nSolution:
\ni) an<\/sub>= 3 + 4n
\na1<\/sub> = 3 + 4(1)
\n= 3 + 4
\n\u2234 an<\/sub> = 7
\n\u2234 a = 7
\na1<\/sub> = 3 + 4n
\na2<\/sub> = 3 + 4 \u00d7 2
\n= 3 + 8
\n\u2234 a2<\/sub> = 11
\nan<\/sub> = 3 + 4n
\na3<\/sub> = 3 + 4 \u00d7 3
\n= 3 + 12
\na3<\/sub> = 15
\n\u2234 a1<\/sub>, a2<\/sub>, a3<\/sub>, ………….
\n7, 11, 15, ……….
\nd = a2<\/sub> – a1<\/sub> = 11 – 7 = 4
\nd = a3<\/sub> – a2<\/sub>= 15 – 11 = 4
\nHere, the value of \u2018d\u2019 is constant.
\n\u2234 an<\/sub> = 3 + 4n forms an Arithmetic Progression.<\/p>\n

ii) an<\/sub> = 9 – 5n
\na1<\/sub>= 9 – 5 \u00d7 1
\n= 9 – 5
\n\u2234 a1<\/sub> = 4
\na1<\/sub> = 9 – 5n
\na1<\/sub> = 9 – 5 \u00d7 2
\n= 9 – 10
\n\u2234 a2<\/sub> = -1
\nan<\/sub>= 9 – 5n
\na3<\/sub> = 9 – 5 \u00d7 3
\n= 9 – 15
\n\u2234 a3<\/sub> = -6
\na1<\/sub>, a2<\/sub>, a3<\/sub>, …………… an<\/sub>
\n4, -1, -6, …………
\nd = a2<\/sub> – a1<\/sub> = -1 – 4 = -5
\nd=a3<\/sub> – a2<\/sub> = -6 – (-1) = -5
\nHere, the value of \u2018d\u2019 is constant.
\n\u2234 an<\/sub> = 9 – 5n form an Arithmetic Progression.<\/p>\n

Arithmetic Progression Class 10 Notes KSEEB Question 11.<\/strong>
\nlf the sum of the first n terms of an AP is 4n – n2<\/sup>, what is the first term (that is SI)? What is the sum of first two term\u0161? What is the second term? Similarly. find the 3rd<\/sup>rd the 10th<\/sup> and the nth<\/sup> terms.
\nSolution:
\nIf Sn<\/sub> = 4n – n2<\/sup>, then
\ni) S1<\/sub> = a =?
\nii) S2<\/sub> =?
\niii) a2<\/sub> =?
\niv) a3<\/sub> =?
\nv) a10<\/sub> =?
\nvi) an<\/sub> =?
\n(i) Sn<\/sub> = 4n – n2<\/sup>
\nS1<\/sub> = 4(1) – 12<\/sup>
\n= 4 – 1
\nS1<\/sub> = 3
\n\u2234 S1<\/sub> = a = 3.<\/p>\n

(ii) S2<\/sub> = 4n – n2<\/sup>
\nS2<\/sub> = 4(2) – 22<\/sup>
\n= 8 – 4
\nS2<\/sub> = 4
\n\u2234 S2<\/sub> = 4<\/p>\n

(iii) We have S2<\/sub> = a1<\/sub> + a2<\/sub> = 4
\n= 3 + a2<\/sub> = 4
\n\u2234 a2<\/sub> = 4 – 3
\n\u2234a2<\/sub> = 1<\/p>\n

(iv) Sn<\/sub> = 4n – n2<\/sup>
\nS3<\/sub>= 4(3) – 32<\/sup>
\n= 12 – 9
\nS3<\/sub> = 3
\na1<\/sub> + a2<\/sub> + a3<\/sub> = 3
\n3 + 1 + a3<\/sub> = 3
\n4 + a3<\/sub>= 3
\n\u2234 a3<\/sub> = 3 – 4
\n\u2234 a 3<\/sub> = -1<\/p>\n

(v) d = a3<\/sub> – a3<\/sub> = -1 -1 = -2
\na10<\/sub>= a + 9d
\n= 3 + 9(-2)
\n= 3 – 18
\na10<\/sub>= -15
\n\u2234 a10<\/sub> = -15<\/p>\n

(vi) an<\/sub> = a + (n – 1)d
\n= 3 + (n – 1)(-2)
\n= 3 + 2n + 2
\n\u2234 an<\/sub> = 5 – 2n
\n\u2234 an<\/sub> = 5 – 2n<\/p>\n

Arithmetic Progression Class 10 KSEEB Solutions Question 12.<\/strong>
\nFind the sum of the first 40 positive integers divisible by 6.
\nSolution:
\n6 + 12 + 18 + 24 + 40 term
\nHere a = 6, d = 2<\/sub> – a1<\/sub> = 12 – 6 = 6
\nn = 40, S40<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{40}=\\frac{40}{2}[2 \\times 6+(40-1) 6]\\)
\n= 20[12 + 39 \u00d7 6]
\n= 20[12 + 234]
\n= 20 \u00d7 246
\n\u2234 S40<\/sub> = 4920.<\/p>\n

Arithmetic Progression 1.3 KSEEB Solutions Question 13.<\/strong>
\nFind the sum of first 15 multiples of 8,
\nSolution:
\nSum of the first 15 multiples of 8 ?
\n8 + 16 + 24+ ……… 15 terms.
\nHere, a = 8, d = a2<\/sub> – a1<\/sub>= 16 – 8 = 8
\nn = 15, S15<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{15}=\\frac{15}{2}[2 \\times 8+(15-1) 8]\\)
\n\\(=\\frac{15}{2}[16+14 \\times 8]\\)
\n\\(=\\frac{15}{2}[16+112]\\)
\n\\(=\\frac{15}{2} \\times 128\\)
\n= 15 \u00d7 64
\n\u2234 S15<\/sub> = 960<\/p>\n

Arithmetic Progression Exercise 1.3 Solution Question 14.<\/strong>
\nFnd the sum of the odd numbers between 0 and 50.
\nSolution:
\nOdd numbers between 0 and 50 are
\n1 + 3 + 5 + 7 + ……… + 49 =?
\nHere. a = 1, d = a3<\/sub> – a1<\/sub>= 3 – 1 = 2
\nan<\/sub> = 49, n =?, S =?
\nan<\/sub> = a + (n – 1) d = 49
\n= 1 + (n – 1)2 = 49
\n1 + 2n – 2 = 49
\n2n – 1 = 49
\n2n = 49 + 1
\n2n = 50
\n\\(n=\\frac{50}{2}\\)
\n\u2234 n = 25
\n\\(S_{n}=\\frac{n}{2}\\left[a+a_{n}\\right]\\)
\n\\(\\mathrm{s}_{25}=\\frac{25}{2}[1+49]\\)
\n\\(=\\frac{25}{2} \\times 50\\)
\n= 25 \u00d7 25
\n\u2234 S25<\/sub> = 625<\/p>\n

Class 10 Maths Chapter 1 Arithmetic Progression KSEEB Solutions Question 15.<\/strong>
\nA contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day. Rs. 250 for the second day. Rs. 300 for the third day. etc.. the penalty for each succeeding day being Rs. 50 more than for the preceeding day. How much money the contractor has to pay as penalty. If he has delayed the work by 30 days?
\nSolution:
\nPenalty for the 1st Day 2nd day 3rd day … 30th<\/sup> day
\nRs. 200 Rs. 250 Rs. 300 Rs.?
\n200 + 250 + 300 + 30 days
\na = 200, d = 250 – 200 = 50, n = 20,
\nS30<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(\\dot{\\mathrm{S}}_{30}=\\frac{30}{2}[2 \\times 200+(30-1) 50]\\)
\n= 15[400 + 29 \u00d7 50]
\n= 15 [400 + 1450]
\n= 15 \u00d7 1850
\n\u2234 S30<\/sub>= Rs. 27750.<\/p>\n

KSEEB Solutions For Class 10 Maths 1.3 Question 16.<\/strong>
\nA sum of Rs. 700 is to be used to give seven cash prizes to students of a school for thler overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
\nSolution:
\nLet the first prize be ‘a’.
\nand Second prize is a – 20 and
\nthe third prize Is a 40.
\na, (a – 20), (a – 40) ……….. n = 7
\na = a, d = a2<\/sub> – a1<\/sub> = a – 20 – a d = -20
\nn = 7, S7<\/sub> =?
\n\\(\\frac{\\mathrm{n}}{2}[2 \\mathrm{a}+(\\mathrm{n}-1) \\mathrm{d}]=\\mathrm{Sn}\\)
\n\\(\\frac{7}{2}[2 \\times a+(7-1)(-20)]=700\\)
\n\\(\\frac{7}{2}[2 a+6(-20)]=700\\)
\n\\(\\frac{7}{2}[2 a-120]=700\\)
\n\\(2 a-120=700 \\times \\frac{2}{7}\\)
\n2a – 120 = 200
\n\u2234 2a = 200 + 120
\n2a = 320
\n\\(\\quad a=\\frac{320}{2}=R s .160\\)
\n\u2234 Each prizes are
\nRs. 160, 140, 120, 100, 80, 60, 40<\/p>\n

Arithmetic Progression KSEEB Solutions Question 17.<\/strong>
\nIn a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant. will be the same as the class. In which they are studying. e.g., a section of Class I will plant 1 tree. a section of Class Il will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
\nSolution:
\n\"KSEEB
\na = 3. d = 6 – 3 = 3, n = 12, S12<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(\\mathrm{S}_{12}=\\frac{12}{2}[2 \\times 3+(12-1) 3]\\)
\n= 6[6 + 11 \u00d7 3]
\n= 6[6 + 33]
\n= 6 \u00d7 39
\n\u2234 S12<\/sub> = 234
\n\u2234 Total number of trees from 3 sections of each class upto 12 class is 234.<\/p>\n

Arithmetic Progression Class 10 Exercise 1.3 KSEEB Solutions Question 18.<\/strong>
\nA spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm. 1.0 cm. 1.5 cm, 2.0 cm as shows In fig. What Is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \\(\\pi=\\frac{22}{7}\\))
\n\"KSEEB
\n[Hint: Length of successive semicircles is l1<\/sub>, l2<\/sub>, l3<\/sub>, l4<\/sub> with centres at A, B, A. B respectively.]
\nSolution :
\n\\(: l_{1}=\\pi \\times \\frac{1}{2}, 12=\\pi \\times 1.13=\\pi \\times \\frac{3}{2}\\)
\n\\(l_{1}=\\frac{\\pi}{2}, \\quad l_{2}=\\mathrm{p}, \\quad l_{3}=\\frac{3}{2} \\pi\\)
\n\u2234 Arithmetic Progression,
\nl1<\/sub>, l2<\/sub>, l3<\/sub>, l4<\/sub>, ………..
\n\"KSEEB<\/p>\n

Arithmetic Progression Class 10 Karnataka State Board KSEEB Solutions Question 19.<\/strong>
\n200 logs are stacked In the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the fig. given below). In how many rows are the 200 logs placed and how many logs are In the top row?
\n\"KSEEB
\nSolution:
\n20, 19, 18, …..
\na = 20, d = 19 – 20 = -1
\nSn<\/sub> = 200, n =?, an<\/sub> =?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(200=\\frac{\\mathrm{n}}{2}[2 \\times 20+(\\mathrm{n}-1)(-1)]\\)
\n\\(200=\\frac{\\mathrm{n}}{2}[40-\\mathrm{n}+1]\\)
\n\\(200=\\frac{n}{2}[41-n]\\)
\n\u2234 400 = n(41 – n)
\n400 = 4n – n2<\/sup>
\n\u2234 n2<\/sup> – 41n + 400 = 0
\nn2<\/sup> – 25n – 16n + 400 = 0
\nn(n – 25) – 16(n – 25) = 0
\n(n – 25) (n – 16) =
\n1f n – 16 = 0 then, n = 16
\n\u2234 an<\/sub> = a + (n – 1) d
\na16<\/sub>= 20 + (16 – 1) (-1)
\n= 20 + 15(-1)
\n= 20 – 15
\n\u2234 a16<\/sub> = 5
\n\u2234 200 logs are placed In 16 rows and there are 5 logs in the top row.<\/p>\n

Exercise 1.3 Arithmetic Progression KSEEB Solutions Question 20.<\/strong>
\nIn a potato race, a bucket is placed at the starting point, which Is 5 m from the first potato. and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see fig, given below)
\n\"KSEEB
\nA competitor starts from the bucket, picks up the nearest potato, runs back with it, drops \u00a1tin the bucket, runs back to pick up the next potato, runs to the bucket to drop It in, and she continues in the same way until al) the potaotes are in the bucket. What is the total distance the competitor has to run?
\n(Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 \u00d7 5 +2 \u00d7 (5 + 3)]
\nSolution:
\nTotal distance competitor taken to pick up the first potato = 5 + 5m
\n= 2 \u00d7 5m.
\n= 10m.
\nTotal distance taken by compeUtor to pick up the second potato
\n= 5 + 3 + 3 + 5m.
\n= 2 \u00d7 5 + 2 \u00d7 3
\n= 2(5 + 3)
\n= 2 \u00d7 8
\n= 16m.
\n\u2234 10m. 16m, 22m 10th potato
\na = 10, d = 16 – 10 = 6m.
\nn = 10, S10 <\/sub>=?
\n\\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\)
\n\\(S_{10}=\\frac{10}{2}[2 \\times 10+(10-1) 6]\\)
\n= 5[20 + 9 \u00d7 6]
\n= 5 [20 + 54]
\n= 5 \u00d7 74
\n\u2234 S10<\/sub> = 370m.
\n\u2234 Total distance the competitor has to run to pick up 10 potatoers is 370 m.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3. Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3 Arithmetic Progression Class 10 Exercise 1.3 …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/215"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=215"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/215\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=215"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=215"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=215"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}