{"id":86,"date":"2021-03-17T11:32:59","date_gmt":"2021-03-17T06:02:59","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=86"},"modified":"2021-03-17T11:32:32","modified_gmt":"2021-03-17T06:02:32","slug":"samacheer-kalvi-10th-maths-chapter-2-ex-2-5","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-10th-maths-chapter-2-ex-2-5\/","title":{"rendered":"Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.5"},"content":{"rendered":"

You can Download Samacheer Kalvi 10th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.5<\/h2>\n

10th Maths Exercise 2.5 Samacheer Kalvi Question 1.<\/strong>
\nCheck whether the following sequences are in A.P.
\n(i) a – 3, a – 5, a – 7, ………
\n(ii) \\(\\frac { 1 }{ 2 } \\),\\(\\frac { 1 }{ 3 } \\),\\(\\frac { 1 }{ 4 } \\),\\(\\frac { 1 }{ 4 } \\), ………..
\n(iii) 9, 13, 17, 21, 25, ………
\n(iv) \\(\\frac { -1 }{ 3 } \\),0,\\(\\frac { 1 }{ 3 } \\),\\(\\frac { 2 }{ 3 } \\),………..
\n(v) 1,-1, 1,-1, 1,-1,…
\nSolution:
\nTo prove it is an A.P, we have to show d = t2<\/sub> – t1<\/sub> = t3<\/sub> – t2<\/sub>.
\n(i) a – 3, a – 5, a – 7………
\nt1<\/sub>,t2<\/sub>,t3<\/sub>
\nd = t2<\/sub> – t1<\/sub> = a – 5 – (a – 3) = a – 5 – a + 3 = -2
\n\u2234 d = -2\u00a0 \u00a0 \u2234 It is an A.P.
\nd = t3<\/sub> – t2<\/sub> = a – 7 – (a – 5) = a – 7 – a + 5 = -2<\/p>\n

(ii) \\(\\frac { 1 }{ 2 } \\),\\(\\frac { 1 }{ 3 } \\),\\(\\frac { 1 }{ 4 } \\),\\(\\frac { 1 }{ 4 } \\), ………..
\n\"10th
\n(v) 1,-1, 1,-1, 1,-1,…
\nd = t2<\/sub> – t1<\/sub> = -1 -1 = -2
\nd = t3<\/sub> – t2<\/sub> = 1 – (-1) = 2
\n-2 \u2260 2 \u2234 It is not an A.P.<\/p>\n

Ex 2.5 Class 10 Samacheer Question 2.<\/strong>
\nFirst term a and common difference d are given below. Find the corresponding A.P.
\n(i) a = 5, d = 6
\n(ii) a = 7, d = 5
\n(iii) a = \\(\\frac { 3 }{ 4 } \\), d = \\(\\frac { 1 }{ 2 } \\)
\nSolution:
\n(i) a = 5, d = 6
\nA.P a, a + d, a + 2d, ………
\n= 5, 5 + 6, 5 + 2 \u00d7 6, ………
\n= 5, 11, 17,…
\n(ii) a = 7,d = -5
\nA.P. = a,a + d,a + 2d,…
\n= 7,7 + (-5), 7 + 2(-5), ……….
\n= 7, 2, -3, …….,…
\n(iii) a = \\(\\frac { 3 }{ 4 } \\), d = \\(\\frac { 1 }{ 2 } \\)
\n\"Ex<\/p>\n

Exercise 2.5 Class 10 Maths Solution Question 3.<\/strong>
\nFind the first term and common difference of the Arithmetic Progressions whose nthterms are given below<\/p>\n

(i) tn<\/sub> = -3 + 2n
\nAnswer:
\ntn<\/sub> = -3 + 2 n
\nt1<\/sub> = -3 + 2(1) = -3 + 2
\n= -1
\nt2<\/sub> = -3 + 2(2) = -3 + 4
\n= 1
\nFirst term (a) = -1 and
\nCommon difference
\n(d) = 1 – (-1) = 1 + 1 = 2<\/p>\n

(ii) tn<\/sub> = 4 – 7n
\nAnswer:
\ntn<\/sub> = 4 – 7n
\nt1<\/sub> = 4 – 7(1)
\n= 4 – 7 = -3
\nt2<\/sub> = 4 – 7(2)
\n= 4 – 14 = -10
\nFirst term (a) = – 3 and
\nCommon difference (d) = 10 – (-3)
\n= – 10 + 3
\n= – 7<\/p>\n

10th Maths Exercise 2.5\u00a0Question 4.<\/strong>
\nFind the 19th term of an A.P. -11, -15, -19, ………..
\nSolution:
\nA.P = -11, -15, -19, ……..
\na = -11
\nd = t2<\/sub> – t1<\/sub> =-15-(-11)
\n= -15 + 11
\n= -4
\nn = 19
\n\u2234 tn<\/sub> = a + (n – 1)d
\nt19<\/sub> = -11 + (19 – 1)(-4)
\n= -11 + 18 \u00d7 -4
\n= -11 – 72
\n= -83<\/p>\n

Class 10 Maths Exercise 2.5 Solution Question 5.<\/strong>
\nWhich term of an A.P. 16, 11, 6, 1,… is -54?
\nAnswer:
\nFirst term (a) = 16
\nCommon difference (d) = 11 – 16 = -5
\ntn<\/sub> = – 54
\na + (n – 1) d = -54
\n16 + (n – 1) (-5) = -54
\n54 + 21 = -54
\n54 + 21 = 5n
\n75 = 5n
\nn = \\(\\frac { 75 }{ 5 } \\) = 15
\nThe 15th<\/sup>\u00a0term is – 54<\/p>\n

Samacheer Kalvi 10th Maths Exercise 2.5 Question 6.<\/strong>
\nFind the middle term(s) of an A.P. 9, 15, 21, 27, ……. ,183.
\nSolution:
\nA.P = 9, 15, 21, 27,…, 183
\nNo. of terms in an A.P. is
\nn = \\(\\frac { l-a }{ d } \\) + 1
\na = 9, l = 183, d = 15 – 9 = 6
\n\u2234 n = \\(\\frac { 183-9 }{ 6 } \\) + 1
\n= \\(\\frac { 174 }{ 6 } \\) + 1
\n= 29 + 1 = 30
\n\u2234 No. of terms = 30. The middle must be 15th term and 16th term.
\n\u2234 t15<\/sub> = a + (n – 1)d
\n= 9 + 14 \u00d7 6
\n=9 + 84
\n= 93
\nt16<\/sub> = a + 15 d
\n= 9 + 15 \u00d7 6
\n= 9 + 90 = 99
\n\u2234 The middle terms are 93, 99.<\/p>\n

10th Maths Exercise 2.5 In Tamil Question 7.<\/strong>
\nIf nine times the ninth term is equal to the fifteen times fifteenth term, Show that six times twenty fourth term is zero.
\nAnswer:
\ntn<\/sub> = a + (n – 1)d
\n9 times 9th<\/sup> term = 15 times 15th<\/sup>\u00a0term
\n9t9<\/sub> = 15 t15<\/sub>
\n9[a + 8d] = 15[a + 14d]
\n9a + 72d = 15a + 210d
\n9a – 15a + 72 d – 210 d = 0
\n-6a – 138 d = 0
\n6a + 138 d = 0
\n6 [a + 23 d] = 0
\n6 [a + (24 – 1)d] = 0
\n6 t24<\/sub> = 0
\n\u2234 Six times 24th terms is 0.<\/p>\n

Ex 2.5 Class 10 Question 8.<\/strong>
\nIf 3 + k, 18 – k, 5k + 1 are in A.P. then find k.
\nSolution:
\n3 + k, 18 – k, 5k + 1 are in A.P
\n\u21d2 2b = a + c if a, b, c are in A.P
\n\"Exercise<\/p>\n

10th Maths 2.5 Exercise Question 9.<\/strong>
\nFind x,y and z gave that the numbers x,
\n10, y, 24, z are in A.P.
\nAnswer:
\nx, 10, y, 24, z are in A.P
\nt2<\/sub> – t1<\/sub> = 10 – x
\nd = 10 – x …..(1)
\nt3<\/sub> – t2<\/sub> = y – 10
\nd = y – 10 ……(2)
\nt4<\/sub> – t3<\/sub> = 24 – y
\nd = 24 – y …..(3)
\nt5<\/sub> – t4<\/sub> = z – 24
\nd = z – 24 …..(4)
\nFrom (2) and (3) we get
\ny – 10 = 24 – y
\n2y = 24 + 10
\n2y = 34
\ny = 17
\nFrom (1) and (2) we get
\n10 – x = y – 10
\n– x – y = -10 -10
\n-x -y = -20
\nx + y = 20
\nx + 17 = 20(y = 17)
\nx = 20 – 17 = 3
\nFrom (1) and (4) we get
\nz – 24 = 10 – x
\nz – 24 = 10 – 3 (x = 3)
\nz – 24 = 7
\nz = 7 + 24
\nz = 31
\nThe value of x = 3, y = 17 and z = 31<\/p>\n

10th Maths 2.5 Question 10.<\/strong>
\nIn a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
\nSolution:
\nt1<\/sub> = a = 20
\nt2<\/sub> = a + 2 = 22
\nt3<\/sub> = a + 2 + 2 = 24 \u21d2 d = 2
\n\u2234 There are 30 rows.
\nt30<\/sub> = a + 29d
\n= 20 + 29 \u00d7 2
\n= 20 + 58
\n= 78
\n\u2234 There will be 78 seats in the last row.<\/p>\n

Exercise 2.5 Class 10 Question 11.<\/strong>
\nThe Sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
\nAnswer:
\nLet the three consecutive terms be a – d, a and a + d
\nBy the given first condition
\na – d + a + a + d = 27
\n3a = 27
\na = \\(\\frac { 27 }{ 3 } \\) = 9
\nAgain by the second condition
\n(a – d) (a) (a + d) = 288
\na (a2<\/sup> – d2<\/sup>) = 288
\n9(81 – d2<\/sup>) = 288 (a = 9)
\n81 – d2<\/sup>\u00a0= \\(\\frac { 288 }{ 9 } \\)
\n81 – d2<\/sup>\u00a0= 32
\n\u2234 d2<\/sup>\u00a0= 81 – 32
\n= 49
\nd = \\(\\sqrt { 49 }\\) = \u00b1 7
\nWhen a = 9, d = 7
\na + d = 9 + 7 = 16
\na = 9
\na – d = 9 – 7 = 2
\nWhen a = 9, d = -7
\na + d = 9 – 7 = 2
\na = 9
\na – d = 9 – (-7) = 9 + 7 = 16
\nThe three terms are 2, 9, 16 (or) 16, 9, 2<\/p>\n

Maths Exercise 2.5 Class 10 Question 12.<\/strong>
\nThe ratio of 6th and 8th<\/sup>\u00a0term of an A.P is 7:9. Find the ratio of 9th<\/sup>\u00a0term to 13th<\/sup>\u00a0term.
\nSolution:
\n\\(\\frac{t_{6}}{t_{8}}=\\frac{7}{9}\\)
\n\\(\\frac { a+5d }{ a+7d } \\) = \\(\\frac { 7 }{ 9 } \\)
\n9a + 45d = 7a + 49d
\n9 a + 45 – 7d = 7a + 49 d
\n9a + 45d – 7a – 49d = 0
\n2a – 4d = 0 \u21d2 2a = 4d
\na = 2d
\nSubstitue a = 2d in
\n\"10th<\/p>\n

Exercise 2.5 Class 10 Maths Samacheer Question 13.<\/strong>
\nIn a winter season let us take the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0\u00b0 C and the sum of the temperatures from Wednesday to Friday is 18\u00b0 C. Find the temperature on each of the five days.
\nSolution:
\nLet the five days temperature be (a – d), a, a + d, a + 2d, a + 3d.
\nThe three days sum = a – d + a + a + d = 0
\n\u21d2 3a = 0 \u21d2 a = 0. (given)
\na + d + a + 2d + a + 3d = 18
\n3a + 6d = 18
\n3(0) + 6 d = 18
\n6d = 18
\nd = \\(\\frac { 18 }{ 6 } \\) = 3
\n\u2234 The temperature of each five days is a – d, a, a + d, a + 2d, a + 3d
\n0 – 3, 0, 0 + 3, 0 + 2(3), 0 + 3(3) = -3\u00b0C, 0\u00b0C, 3\u00b0C, 6\u00b0C, 9\u00b0C<\/p>\n

2.5 Exercise Class 10 Question 14.<\/strong>
\nPriya earned \u20b9 15,000 in the first month. Thereafter her salary increased by \u20b91500 per year. Her expenses are \u20b913,000 during the first year and the expenses increases by \u20b9900 per year. How long will it take for her to save \u20b920,000 per month.
\nSolution:
\n\"Class
\nWe find that the yearly savings is in A.P with a1<\/sub> = 2000 and d = 600.
\nWe are required to find how many years are required to save 20,000 a year …………..
\nan<\/sub> = 20,000
\nan<\/sub> = a + (n – 1)d
\n20000 = 2000 + (n – 1)600
\n(n – 1)600 = 18000
\nn – 1 = \\(\\frac { 18000 }{ 600 } \\) = 30
\nn = 31 years<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.5 10th Maths Exercise 2.5 Samacheer Kalvi Question 1. Check whether the following sequences are …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/86"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=86"}],"version-history":[{"count":0,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/86\/revisions"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=86"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=86"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=86"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}