{"id":24989,"date":"2020-03-04T15:01:51","date_gmt":"2020-03-04T09:31:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=24989"},"modified":"2021-07-02T15:21:21","modified_gmt":"2021-07-02T09:51:21","slug":"kseeb-solutions-for-class-10-maths-chapter-1-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-1-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions<\/h2>\n

I. Multiple-choice Questions:<\/span><\/p>\n

Question 1.
\nIf an<\/sub> = n2<\/sup> – 1 and an<\/sub> = 99 then the value of n is
\na) 100
\nb) 10
\nc) 9
\nd) 99
\nAnswer:
\nb) 10<\/p>\n

Question 2.
\nIf an<\/sub>= 2n2<\/sup> – 1 then the value of 4th<\/sup> term is
\na) 32
\nb) 30
\nc) 31
\nd) 18
\nAnswer:
\nc) 31<\/p>\n

Question 3.
\nIf an<\/sub>= 2n + 1 then the common difference of the AP is
\na) 3
\nb) 5
\nc) 2
\nd) 1
\nAnswer:
\nc) 2<\/p>\n

Question 4.
\nThe general form of an A.P is
\na) a, ar, ar2<\/sup>,…
\nb) a, a + d, a + 2d,…
\nc) a, a-d,.a-2d,….
\nd) \\(\\frac{1}{a}, \\frac{1}{a+d}, \\frac{1}{a+2 d}\\) …..
\nAnswer:
\nb) a, a + d, a + 2d,…<\/p>\n

Question 5.
\nIn an A.P the common difference is 3 first term is 1 then the value of 10th<\/sup> term is
\na) 28
\nb) 27
\nc) 25
\nd) 40
\nAnswer:
\na) 28<\/p>\n

Question 6.
\nThe formula to find nth<\/sup> term of an A.P is.
\na) a + (n – 1)d
\nb) arn-1<\/sup>
\nc) \\(\\frac{1}{a+(n-1) d}\\)
\nd) a – (n + 1)d
\nAnswer:
\na) a + (n – 1)d<\/p>\n

Question 7.
\nIf 20, x + 1, 4 are In AP then value of x is
\na) 11
\nb) 12
\ne) 24
\nd) 13
\nAnswer:
\na) 11<\/p>\n

Question 8.
\nIn an arithmetic progression an+5<\/sub> = 35 & an+1<\/sub> = 23 then common difference is
\na) 3n
\nb) 4n
\nc) 2
\nd) 3
\nAnswer:
\nd) 3<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nIn an A.P the relation between a5<\/sub> and a7<\/sub> and common difference is (d) is
\na) a5<\/sub> = a7<\/sub> + 2d
\nb) a5<\/sub> = a7<\/sub> + d
\nc) a7<\/sub> = a5<\/sub> + 3d
\nd) a7<\/sub> = a5<\/sub> + 2d
\nAnswer:
\nd) a7 = a5 + 2d<\/p>\n

Question 10.
\nIf a is constant then a + 2a + 3a +…. + na is
\n\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 11.
\nThe value of \u03a3(n – 1) s
\na) \\(\\frac{\\mathrm{n}(\\mathrm{n}+1)}{2}\\)
\nb) \\(\\frac{n(n-1)}{2}\\)
\nc) \\(\\frac{n}{2}\\)
\nd) \\(\\frac{n^{2}}{2}\\)
\nAnswer:
\nb) \\(\\frac{n(n-1)}{2}\\)<\/p>\n

Question 12.
\nIn an A.P, if S5<\/sub> = 35 and S4<\/sub> = 22 then 5th<\/sup> term is.
\na) 35
\nb) 10
\nc) 13
\nd) 22
\nAnswer:
\nc) 13<\/p>\n

Question 13.
\nThe nth<\/sup> term of 3, 7, 11, 15, is
\na) 4n – 1
\nb) 4n + 1
\nc) 4n + 3
\nd) 3n + 4
\nAnswer:
\na) 4n – 1<\/p>\n

Question 14.
\nIn an A.P, if a4<\/sub> = 8 & a = 2, then its common difference is
\na) 6
\nb) 4
\nc) 2
\nd) 10
\nAnswer:
\nc) 2<\/p>\n

Question 15.
\nIn an A.P an+5<\/sub> = 50 and an+1<\/sub>\u00a0= 38 then common difference
\na) 3
\nb) 2
\nc) 3n
\nd) 2n
\nAnswer:
\na) 3<\/p>\n

Question 16.
\nAmong the following arithmetic progression is
\na) 1, 4, 6,…
\nb) 12, 10, 14,…
\nc) 35, 32, 25,….
\nd) 8, 13, 19,
\nAnswer:
\nc) 35, 32, 25,….<\/p>\n

Question 17.
\nIn an AP, the correct relation is
\na) an-5<\/sub> = an-4<\/sub>\u00a0+ d
\nb) an-5<\/sub> = an-6<\/sub>\u00a0+ d
\nC) an-5<\/sub> = an<\/sub>\u00a0+ d
\nd) an+5<\/sub> = an-5<\/sub> – d
\nAnswer:
\nb) an-5<\/sub> = an-6 <\/sub>+ d<\/p>\n

Question 18.
\nThe value of \\(\\sum_{n=1}^{n}10\\) is
\na) 10
\nb) 11
\nc) 55
\nd) 110
\nAnswer:
\nc) 55<\/p>\n

Question 19.
\nIf 2x + 1, 4x, 13-x are in Arithmetic progression then x is equal to
\na) 2
\nb) 3
\nc) 4
\nd) 5
\nAnswer:
\na) 2<\/p>\n

Question 20.
\nIn a progression, if an<\/sub> = 2n2<\/sup> + 1, then S2 <\/sub>is
\na) 9
\nb) 12
\nc) 10
\nd) 11
\nAnswer:
\nb) 12<\/p>\n

II. Short answer Questions:<\/span><\/p>\n

Question 1.
\nWrite the formula to find nth<\/sup> term of an A.P?
\nAnswer:
\nan<\/sub> = a + (n – 1)d<\/p>\n

Question 2.
\nWrite the formula to find arithmetic mean of 3 numbers
\nAnswer:
\nAM = \\(\\frac{a+b+c}{3}\\)<\/p>\n

Question 3.
\nThe first term of an A.P is q and its common difference is p. Find its 20th term.
\nAnswer:
\na20<\/sub> = a + 19d
\n= q + 19p<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIf an<\/sub> = 5n<\/sub> – 3, find the common difference
\nAnswer:
\nan<\/sub> = 5n – 3
\na1<\/sub> = 5(1) – 3 = 5 – 3 = 2
\na2<\/sub>\u00a0= 5(2) – 3 = 10 – 3 = 7
\nd = a2<\/sub> – a1<\/sub> = 7 – 2 = 5
\ncommon difference = 5<\/p>\n

Question 5.
\nFind the 9th term from the end (towards the first term) of the A.P 5, 9, 13, ….185
\nAnswer:
\na = 185, d = – 4 & 1 = 5
\na9<\/sub> = a + (n – 1)d
\na9<\/sub>= 1 + (9 – 1) – 4
\na9<\/sub> = 185 + 8x – 4
\n= 185 – 32
\na9<\/sub> = 153<\/p>\n

Question 6.
\nFor what value of p are 2p + 1, 13, 5p – 3, three consecutive terms of a AP?
\nAnswer:
\na2<\/sub> – a1<\/sub>, = a3<\/sub> – a2<\/sub>
\n13 – (2p + 1) = 5p – 3 – 13
\n13 – 2p – 1 = 5p – 16
\n12 + 16 = 5p + 2p
\n7p = 28
\n\"KSEEB
\np = 4<\/p>\n

Question 7.
\nFind the common difference of an AP in which a18<\/sub> – a14<\/sub> = 32
\nAnswer:
\na18<\/sub> – a14<\/sub> = 32
\n(a + 17d) – (a + 13d) = 32
\na + 17d – a – 13d = 32
\n4d = 32
\n\"KSEEB
\nd = 8<\/p>\n

Question 8.
\nIn an A.P an<\/sub> = 20 & Sn<\/sub> = 399 find Sn-1<\/sub>
\nAnswer:
\nan<\/sub> = Sn<\/sub> – Sn-1<\/sub>
\nSn-1<\/sub> = Sn<\/sub> – an<\/sub>
\n= 399 – 20
\nSn-1<\/sub> = 379<\/p>\n

Question 9.
\nIn an AP, if d = – 4, n = 7, an<\/sub> = 4 then find a.
\nAnswer:
\nan<\/sub> = a + (n – 1)d
\n4 = a + (7 – 1) – 4
\n4 = a + 6 x – 4
\n4 = a – 24
\na = 4 + 24
\na = 28<\/p>\n

III. Long Answer Questions:<\/span><\/p>\n

Question 1.
\nFind the number of terms in the A.P, 100, 96, 92,….., 12
\nAnswer:
\n100, 96, 92,………, 12
\na = 100 d = a2<\/sub> – a1<\/sub>, d = 96 – 100 = – 4
\nan<\/sub> = 12 & n = ?
\nan<\/sub> = a + (n – 1)d
\n12 = 100 + (n – 1)d
\n12 – 100 = (n – 1) x – 4
\n– 88 = (n – 1) x – 4
\n\\(\\frac{-88}{-4}\\) = n – 1
\nn – 1 = 22
\nn = 22 + 1
\nn = 23<\/p>\n

Question 2.
\nThe angles of a triangle are in A.P if the smallest angle is 50\u00b0, find the other two angles.
\nAnswer:
\nLet the smallest angle = 50\u00b0
\nThree angles of AP are a, a + d, a + 2d – Sum of the angles of a triangle = 180\u00b0 a + a + d + a + 2d = 180\u00b0
\n50\u00b0 + 50\u00b0 + d + 50\u00b0 + 2d = 180\u00b0
\n3d = 180\u00b0- 150\u00b0
\n3d = 30\u00b0
\n\"KSEEB
\nd = 10\u00b0<\/p>\n

Required angles a = 50\u00b0
\na + d = 50\u00b0 + 10\u00b0 = 60\u00b0
\na + 2d = 50 + 2(10) = 50 + 20 = 70\u00b0<\/p>\n

Question 3.
\nAn A.P consists of 50 terms of which 3rd<\/sup> term is 12 and last term is 106. Find the 29th<\/sup> term.
\nAnswer:
\nn = 50
\na3<\/sub> = 12
\na + 2d = 12
\na = 12 – 12d \u2192 (1)
\na = 106
\na29<\/sub> = ?
\nan<\/sub> = a + (n – 1) d
\n106 = 12 – 2d + (50 – 1)d
\n106 – 12 = – 2d + 49d
\n94 = 47d
\n\"KSEEB
\nd = 2
\nPut d = 2 is eqn (1)
\na = 12 – 2 (2)
\na = 12 – 4
\na = 8<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nThe sum of 4th<\/sup> and 8th<\/sup> terms of an A.P is 24 and the sum of 6th<\/sup> and 10th<\/sup> terms of the same A.P is 44. Find the first three terms.
\nAnswer:
\nSum of 4th<\/sup> and 8th<\/sup> terms = 24
\na4<\/sub> + a8<\/sub> = 24
\na + 3d + a + 7d = 24
\n2a + 10d = 24
\n\u00f7 by 2 on both the sides
\na + 5d = 12 \u2192 (1)
\nSum of 6th<\/sup> and 10th<\/sup> terms = 44
\na6<\/sub> + a10<\/sub> = 44
\na + 5d + a + 9d = 44
\n2a + 14d = 44 by 2
\na + 7d = 22 \u2192 (2)
\nSubtract eqn (1) & (2)
\na + 5d = 12
\na + 7d = 22
\n(-) (-) (-)
\n________
\n– 2d = – 10
\n+ d = \\(\\frac{10}{2}\\) = 5<\/p>\n

Put d = 5 is eqn (1)
\na = 12 – 5d
\na = 12 – 5(5)
\n= 12 – 25 a = – 13
\nrequired terms a, a + d, a + 2d…….
\n– 13, – 13,+ 5, – 13 + 2(5)
\n– 13,- 8, – 3<\/p>\n

Question 5.
\nThe ratio of 7th<\/sup> to 3rd<\/sup> term of an A.P is 12 : 5 find the ratio of 13th<\/sup> to 4th<\/sup> term
\nAnswer:
\n\\(\\frac{a_{7}}{a_{3}}=\\frac{12}{5}\\)
\n\\(\\frac{a+6 d}{a+2 d}=\\frac{12}{5}\\)
\n5(a + 6d) = 12 (a + 2d)
\n5a + 30d= 12a + 24d
\n30d – 24d = 12a – 5a
\n6d = 7a
\nd = \\(\\frac{7}{6}\\) a
\nThe ratio of 13th<\/sup> to 4th<\/sup> term
\n\"KSEEB<\/p>\n

Question 6.
\nA company employed 400 persons in the year 2001 and each year increased by 35 persons. In which year the number of employees in the company will be 785?
\nAnswer:
\n2001 + 2002 ………………..
\n400, 435, 470, ………….. 785
\na = 400, d = a2<\/sub> – a1<\/sub> an = 785, n = ? d = 435 – 400
\nd = 35
\nan<\/sub> = a + (n – 1)d
\n785 = 400 + (n – 1) 35
\n785 – 400 = (n – 1) \u00d7 35
\n385 = (n – 1) x 35
\nn – 1 = \\(\\frac{385}{35}\\)
\nn – 1 = 11
\nn = 11 + 1
\nn = 12
\nEmployes in the company will be 785 reach in year 2012.<\/p>\n

Question 7.
\nIf the pth term of an A.P is q and the qth term is p. Prove that the nth<\/sup> term equal to (p + q – n)
\nAnswer:
\nLet a be the first term and ‘d’ the common difference
\nap<\/sub> = a + (p – 1)d and aq<\/sub> = a + (q – 1)d
\nNow ap<\/sub> = q and aq<\/sub> = p given
\na + (p – 1)d = q \u2192 (1)
\na + (q – 1)d = p \u2192 (2)
\nOn subtracting (1) from (2) we get
\n\"KSEEB<\/p>\n

Putting d = – 1 in the eqn (i) we get a + (p – 1 ) x – 1 = q
\na – p + 1 = q a = p + q – 1
\n\u2234 nth<\/sup> term = a + (n – 1)d
\n= p + q – 1 + (n – 1) x – 1
\n= p + q – 1 + n + 1
\nnth<\/sup> term = p + q – n
\nHence it is proved<\/p>\n

Question 8.
\nFind four numbers in A.P such that the sum of 2nd<\/sup> and 3rd<\/sup> terms is 22 and the product of 1st<\/sup> and 4th<\/sup> terms is 85.
\nAnswer:
\nSum of 2nd<\/sup> and 3rd<\/sup> terms = 22
\n\"KSEEB<\/p>\n

Question 9.
\nFind the sum of all natural numbers between 1 and 201 which are divisible by 5.
\nAnswer:
\n5 + 10 +15 +………….+ 200
\nwhich are divisible by 5
\na = 5
\nd = a2<\/sub> – a1<\/sub> = 10 – 5 = 5
\nl = an<\/sub> = 200 + (n – 1)d
\n200 = 5 + (n – 1) x 5
\n195 = (n – 1) x 5
\nn – 1 = \\(\\frac{195}{5}\\)
\n\"KSEEB<\/p>\n

Question 10.
\nFind three numbers in AP whose sum and products are respectively. 21 and 231
\nAnswer:
\nLet the numbers be a – d, a, a + d
\nSum = 21
\n\"KSEEB<\/p>\n

Question 11.
\nThe sum of 6 terms which form an A.P is 345. The difference b\/w the first and last terms is 55 find the terms.
\nAnswer:
\nLet the numbers be a, a + d, a + 2d, a + 3d
\na + 4d, a + 5d, a + 6d
\n6d + 15d = 345 – 3
\n2a + 5d = 115 \u2192 (1)
\nthe difference b\/w first term and last term is 55
\na6<\/sub> – a1<\/sub> = 55
\na + 5d – a = 55
\n5d = 5
\nd = \\(\\frac{55}{5}\\)
\nd = 11
\nPut d = 11 in eqn (1)
\n2a + 5d = 115
\n2a + 55 = 115
\n2a = 115 – 55
\n2a = 60
\na = \\(\\frac{60}{2}\\)
\na = 30
\nrequired numbers
\na, a + d, a + 2d, a + 3d, a + 4d, a + 5d
\n30, 30 + 11, 30 + 2(11), 30 + 3(11), 30 + 4(11), 30 + 5(11)
\n30, 41, 52, 63, 74, 85.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nIn an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms show that T20<\/sub> = – 112, Find S20<\/sub>.
\nAnswer:
\nLet the five number a, a + d, a + 2d, a + 3d, a + 4d, next 5 numbers be
\na + 5d, a + 6d, a + 7d, a + 8d, a + 9d a = 2
\na + a + d + a + 2d + a + 3d + a + 4d
\n= \\(\\frac{1}{4}\\) (a + 5d + a + 6d + a + 7d + a + 8d + a + 9d)
\n5a + 10d = \\(\\frac{1}{4}\\) (5a + 35d)
\n5(2) + 10d = \\(\\frac{1}{4}\\) [(5 x 2) + 35d]<\/p>\n

\"KSEEB
\n4(10 + 10d) = 10 + 35d
\n40 + 40d = 10 + 35d
\n40d – 35d = 10 – 40
\n5d = – 30
\nd = \\(\\frac{-30}{5}\\)
\nd = – 6
\na20<\/sub> = ?, n = 20
\nan<\/sub> = a + (n – 1)d
\na20<\/sub> = 2 + (20 – 1) x 6
\n= 2 – 114 a20<\/sub> = – 112
\na20<\/sub> = l = – 112
\nS20<\/sub> = ? n = 20
\nS = f{2+(-112)]
\ns20<\/sub> =10(2-112) = 10(-110)
\ns20<\/sub> = -1100<\/p>\n

Question 13.
\nThe third term of an A.P is 8 and the ninth term of the A.P exceeds three times the third term by 2 find the sum of its first 19 terms.
\nAnswer:
\na3<\/sub> = 8 (given)
\na + 2d = 8 \u2192 (1)
\na9<\/sub> = 3(a3<\/sub>) + 2 (given)
\na + 8d = 3(8) + 2 [bye qn (1)]
\na + 8d = 26
\na + 2d + 6d = 26
\n8 + 6d = 26 (a + 2d = 28)
\n6d = 26 – 8
\n6d = 18
\nd = \\(\\frac{18}{6}\\)
\nd = 3
\nPut d = 3 in eqn (1)
\na + 2(3) = 8
\na + 6 = 8
\na = 8 – 6
\na = 2
\nn = 19
\n\"KSEEB<\/p>\n

Question 14.
\nIf the sum of m terms of an AP is the same as the sum of its n terms, show that the sum of its (m + n) terms is 120\u00b0. Find the number of sides of the zero.
\nAnswer:
\nLet a be the first term and d be the common difference of the given AP
\nThen Sm<\/sub> = Sn<\/sub>
\n\"KSEEB
\nQuestion 15.
\nFind the sum of all the first \u2018n\u2019 odd natural numbers.
\nAnswer:
\n1, 3, 5, 7, 9,…… up to n
\na = 1, d = 3 – 1 = 2, n = n
\n\"KSEEB<\/p>\n

Question 16.
\nThe common difference between any two consecutive Interior angles of a polygon is 5\u00b0. If the smallest angle is 120\u00b0. Find the number of sides of the polygon?
\nAnswer:
\nLet the AP be 120\u00b0, 125\u00b0…… with
\na = 120\u00b0and d = 5
\nThe number of sides in a polygon is the sum of its interior angles. Therefore (2n – 4)90\u00b0
\n= 180n – 360\u00b0
\n\"KSEEB
\n2 [180n- 360] = n [240 + 5n – 5]
\n360n – 720 = n (235 + 5n)
\n360n – 720 = 235n + 5n2<\/sup>
\n5n2<\/sup> + 235n – 360n + 720 = 0
\n5n2<\/sup> – 125n + 720 = 0 + by 5
\nn2<\/sup>-25n + 144 = 0
\nn2<\/sup> – 16n – 9n + 144 0
\nn(n – 16) – 9(n – 16) = 0
\n(n – 16) (n – 9) = 0
\nn – 16 = 0, and n – 9 = 0
\nn = 16 and n = 9
\n\u2234 n = 9
\n\u2234 number of sides is 9 so it is a nanogon.<\/p>\n

Question 17.
\nIn an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms show that T20<\/sub> = – 112
\nAnswer:
\na = 2,\u00a0 S5<\/sub> =\\(\\frac{1}{4}\\)[S10 <\/sub>– S5<\/sub>], T20<\/sub> = – 112
\n\"KSEEB<\/p>\n

Question 18.
\nIn an arithmetic progression the sum of first 10 terms is 175 and the sum of the next 10 terms is 475 find the arithmetic progression.
\nAnswer:
\n\"KSEEB
\n\"KSEEB
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Additional Questions I. Multiple-choice Questions: Question 1. …<\/p>\n","protected":false},"author":5,"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\/24989"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=24989"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/24989\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=24989"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=24989"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=24989"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}