{"id":152,"date":"2021-07-24T14:00:52","date_gmt":"2021-07-24T08:30:52","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=152"},"modified":"2021-07-24T14:18:47","modified_gmt":"2021-07-24T08:48:47","slug":"kseeb-sslc-class-10-maths-solutions-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-1-ex-1-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 <\/strong>are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2<\/h2>\n

Arithmetic Progression Class 10 Exercise 1.2 Solutions Question 1.<\/strong>
\nFill in the blanks In the following table,
\ngiven that ‘a’ is the first term, ‘d’ is the common difference and an<\/sub> the nth<\/sup> term of the A.P.
\n\"KSEEB
\nSolution:
\n(i) a = 7, d = 3, n = 8, an<\/sub> =?
\nan<\/sub> = a + (n – 1) d
\na8<\/sub>= 7 + (8 – 1) 3
\n= 7 + 7 \u00d7 3
\n= 7 + 21
\n\u2234 a8<\/sub> = 28<\/p>\n

(ii) a = -18, d =?, n = 10, an<\/sub> = 0
\nan<\/sub> = a + (n – 1) d
\n0 = -18 + (10 – 1) d
\n0 = -18 + 9d
\n18 = 9d
\n9d = 18
\n\\(\\mathrm{d}=\\frac{18}{9}=2\\)<\/p>\n

(iii) a =?, d = -3, n = 18, an<\/sub> = -5
\nan<\/sub> = a + (n – 1) d
\n-5 = a + (18 – 1) (-3)
\n= a + 17(-3)
\n-6 = 1 – 51
\n\u2234 a = -5 + 51 = 46<\/p>\n

(iv) a = -18.9, d = 2.5, n =? an<\/sub> = 3.6
\nan<\/sub> = a + (n – 1) d
\n3.6= -18.9 + (n – 1) (2.5)
\n3.6= -18.9 + 2.5n – 2.5
\n3.6 = 2.5n – 21.4
\n2.5n = 3.6 + 21.4
\n2.5n = 25
\n\\(n=\\frac{25}{2.5}=\\frac{250}{25}\\)
\n\u2234 n = 10.<\/p>\n

(v) a = 3.5, d = 0, n = 105, an<\/sub> =?
\nan<\/sub> = a + (n – 1) d
\n= 3.5 + (105 – 1) (0)
\n= 3.5+ 104 \u00d7 0
\n= 3.5 +0
\n\u2234 an<\/sub> = 3.5<\/p>\n

Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 2.<\/strong>
\nChoose the correct choice In the following and justify:
\n(i) 30th<\/sup> term of the AP: 10, 7, 4, ……….. is
\nA) 97
\nB) 77
\nC) -77
\nD) -87
\nSolution:
\na = 10. d = 7 – 10 = -3, n = 30, a30<\/sub> =?
\nan<\/sub> = a + (n – 1) d
\na30<\/sub> = 10 + (30 – 1)(-3)
\n= 10 + 29(-3)
\n= 10 – 87
\n\u2234 a30<\/sub> = 77
\n\u2234 Ans: (C) -77<\/p>\n

(ii) 11th<\/sup> term of the A.P.
\n\\(-3,-\\frac{1}{2}, 2, \\dots \\ldots,\\) is
\nA) 28
\nB) 22
\nC) -38
\nD) \\(-48 \\frac{1}{2}\\)
\nSolution:
\n\\(a=-3, \\quad d=-\\frac{1}{2}-(-3)=-\\frac{1}{2}+3=2 \\frac{1}{2}\\)
\nn = 11, a11<\/sub> =?
\nan<\/sub> = a + (n – 1) d
\n\\(a_{11}=-3+(11-1)\\left(2 \\frac{1}{2}\\right)\\)
\n\\(=-3+10\\left(\\frac{5}{2}\\right)\\)
\n= -3 +25
\n\u2234 a11<\/sub> = 22
\n\u2234 Ans: (B) 22<\/p>\n

KSEEB Solutions For Class 10 Maths Arithmetic Progression Exercise 1.2 Question 3.<\/strong>
\nIn the following APs find the missing terms in the boxes:
\n\"KSEEB
\nSolution:
\n(i) \"KSEEB
\na = 2, a + d =?, a + 2d = 26
\na + 26 = 26
\n2 + 2d = 26
\n2d = 24
\n\u2234 d=12 . .
\n\u2234 a + d = 2 + 12 = 14
\n\u2234 Ans: 14<\/p>\n

(ii) Here, a =?, a + d = 13, a + 2d=?,
\na + 3d = 3
\na + d + 2d = 3
\n13 + 2d = 3
\n2d = 3 – 13
\n2d = -10
\n\u2234 d = -5
\na + d = 13
\na + (-5) = 13
\na – 5 = 13
\n\u2234 a – 13 + 5 = 18
\n\u2234 a = 18
\na + 2d =?
\n= 18 + 2(-5)
\n= 18 – 10
\na + 2d = 8
\n\u2234 Ans: 18, 8<\/p>\n

(iii) a = 5, a + d =?, a + 2d =?
\n\"KSEEB<\/p>\n

(iv) a = -4, a +d=? a + 2d =?
\na+3d = ? a + 4d =? a + 5d = 6
\na + 5d = 6
\n-4 + 5d = 6
\n5d = 6 + 4
\n5d = 10
\n\\(\\mathrm{d}=\\frac{10}{5}\\)
\n\u2234 d=2.
\na + d = -4 + 2 = -2
\na + 2d = -4 + 2 (2) = -4 + 4 = 0
\na + 3d = -4 + 3 (2) = -4 + 6 = 2
\na + 4d = -4 + 4 (2) = -4 + 8 = 4
\n\u2234 Ans: -2, 0, 2, 4<\/p>\n

(v) a =? a + d = 38, a + 2d =?
\na + 3d =?, a + 4d =?a + 5d = 22
\n\"KSEEB
\n4d = -60
\n\\(\\mathrm{d}=-\\frac{-60}{4} \\quad=-15\\)
\na + d = 38
\na – 15=38
\na = 38 + 15 = 53
\na + 2d = 53 + 2(-15) = 53 – 30 = 23
\na + 3d = 53 + 3(-15) = 53 – 45 = 8
\na + 4d =53 + 4(-15) = 53 – 60 = 7
\n\u2234 53, 23, 8, -7.<\/p>\n

Exercise 1.2 Class 10 Maths Arithmetic Progression KSEEB Solutions Question 4.<\/strong>
\nWhich term of the AP : 3, 8, 13. 18, ………. is 78?
\nSolution:
\na = 3, d = 8 – 3 = 5, an<\/sub> = 78, n =?
\nan<\/sub> = a + (n – 1) d
\n78 = 3+(n\u20141)(5)
\n78 = 3 + 5n – 5
\n78 = 5n – 2
\n5n = 78 + 2
\n\\(n=\\frac{80}{5}\\)
\n\u2234 n = 16<\/p>\n

10th Maths Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 5.<\/strong>
\nFind the number of terms In each of the following APs:
\n(i) 7, 13, 19, ………… 201
\n(ii) \\(18,15 \\frac{1}{2}, 13, \\dots \\dots 47\\)
\nSolution:
\n(i) 7, 13, 19, ………. 201
\na = 3, d = 13 – 7 = 6, an<\/sub> = 201. n =?
\na + (n – 1) d = an<\/sub>
\n3 + (n – 1) 6 = 201
\n3 + 6n – 6 = 201
\n6n – 3 = 201
\n6n = 203
\n\\(n=\\frac{204}{6}\\)
\n\u2234 n = 34<\/p>\n

(ii) \\(18,15 \\frac{1}{2}, 13, \\dots \\dots 47\\)
\n\\(a=18, d=a_{2}-a_{1}=\\frac{31-36}{2}=\\frac{-5}{2}\\)
\nan<\/sub> = -47, n =?
\nan<\/sub> = a + (n – 1) d
\n\\(-47=18+(n-1)\\left(\\frac{-5}{2}\\right)\\)
\n\\(-47-18=(n-1)\\left(\\frac{-5}{2}\\right)\\)
\n\\((n-1)\\left(\\frac{-5}{2}\\right)=-65\\)
\n\\(n-1=-65 \\times \\frac{-2}{5}\\)
\nn – 1 = -13 \u00d7 -2
\nn – 1 = + 26
\n\u2234 n = 26 + 1
\n\u2234 n = 27<\/p>\n

KSEEB Solutions For Class 10 Maths Arithmetic Progression Question 6.<\/strong>
\nCheck whether -150 is a term of the A.P: 11, 8, 5, 2, ………..
\nSolution:
\n11, 8, 5, 2, ……….. -150
\na = 11, d = 8 11 = -3. an<\/sub> = -150.
\na + (n – 1) d = an<\/sub>
\n11 + (n – 1) (-3) = -150
\n11 – 3n + 3 = -150
\n-3n + 14 = -150
\n-3n = -150 – 14
\n-3n = -164
\n3n = 164
\n\\(n=\\frac{164}{3}\\)
\nHere value of \u2018n\u2019 is not perfect. Hence -150 is not a term of the A.P.<\/p>\n

Arithmetic Progression Class 10 Exercise 1.2 KSEEB Solutions Question 7.<\/strong>
\nFind the 31st<\/sup> term of an AP whose 11th<\/sup> term Is 38 and the 16th<\/sup> term is 73.
\nSolution:
\na = 38, a16<\/sub> = 83 a31<\/sub> =?
\nan<\/sub> = a + (n – 1) d
\na16<\/sub> = a + (16 – 1) d
\na + 15d = 83
\n38 + 15d = 83
\n15d = 83 – 38
\n15d = 45
\n\\(d=\\frac{45}{15}\\)
\n\u2234 d=3.
\n\u2234 an<\/sub> = a + (n – 1)d
\na31<\/sub> = 38 + (31 – 1) 3
\n= 38 + 30 \u00d7 3
\n= 38 + 90
\n\u2234 a = 128.<\/p>\n

Arithmetic Progression Class 10 KSEEB Solutions Question 8.<\/strong>
\nAn AP consists of 50 terms of which 3rd<\/sup> term is 12 and the last term is 106. Find the 29th<\/sup> term.
\nSolution:
\nn = 50, a3<\/sub> = 12, an<\/sub> = 106, a29<\/sub> =?
\na11<\/sub> = a + (n – 1) d
\na50<\/sub> = a + (50 – 1) d = 106
\n\u2234 a + 49d = 106 ……………… (1)
\na3 <\/sub>= a + 2d = 12 ………………. (2)
\nSubtracting equation (2) in equation (1).
\n\"KSEEB
\nSubstituting the value of d.
\na + 2d = 12
\na + 2(2) = 12
\na + 4 = 12
\n\u2234 a = 12 – 4
\na = 8.
\n\u2234 an<\/sub> = a + (n – 1) d
\na29<\/sub> = 8 + (29 – 1) 2
\n= 8 + 28 \u00d7 2
\n= 8 + 56
\n\u2234 a29<\/sub> = 64<\/p>\n

10th Standard Maths Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 9.<\/strong>
\nIf the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
\nSolution:
\na3<\/sub> = 4, a9<\/sub> = -8, an<\/sub> = 0, n =?
\na3<\/sub> = a + 2d = 4 ………….. (1)
\na9<\/sub> = a + 8d = -8 …………. (2)
\nFrom equation (1) – equation (2).
\n\"KSEEB
\n6d = 12
\n\\({d}=\\frac{-12}{6}\\)
\n\u2234 d = -2
\na + 2d = 4
\na – 2(2) = 4
\na – 4 =4
\n\u2234 a = 4 + 4
\n\u2234 a = 8
\nan<\/sub> = a + (n – 1) d
\n= 8 + (n – 1) (-2)
\n= 8 – 2n + 2
\n= 10 – 2n = 0 \u2235 an<\/sub> = 0
\n\\(n=\\frac{10}{2}\\)
\n\u2234 n = 5
\n\u2234 5th<\/sup> term of this AP is Zero.<\/p>\n

Maths Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 10.<\/strong>
\nThe 1 7th<\/sup> term of an AP exceeds its 17th<\/sup> term by 7. Find the common difference.
\nSolution:
\na17<\/sub> = a10<\/sub> + 7, d =?
\na + 16d = a + 9d + 7
\na +1 6d – a – 9d = 7
\n7d = 7
\n\\(\\mathrm{d}=\\frac{7}{7}\\)
\n\u2234 = 1<\/p>\n

10th Arithmetic Progression Exercise 1.2 KSEEB Solutions Question 11.<\/strong>
\nWhich term of the AP: 3, 15, 27. 39, … will be 132 more than its 54th<\/sup> term?
\nSolution:
\n3, 15, 27, 39, ………… an<\/sub> =?, n =?
\nan<\/sub> = a54<\/sub> + 132
\na = 3, d = 15 – 3 = 12
\nan<\/sub> = a54<\/sub> + 132
\nan<\/sub> = a + 53d + 132
\n3 + 53(12) + 132
\n= 3 + 636 + 132
\n\u2234 an<\/sub> = 771
\nan<\/sub> = a + (n – 1) d = 771
\n= 3 + (n – 1)12 = 771
\n3 + 12n \u2014 12 = 771
\n12n – 9 = 771
\n12n = 771 + 9
\n12n = 780
\n\\(n=\\frac{780}{12}\\)
\n\u2234 n = 65.
\n\u2234 65th<\/sup> term is 132 more than its 54th term.<\/p>\n

Arithmetic Progression Class 10 Notes KSEEB Question 12.<\/strong>
\nTwo APs have the same common difference. The difference between their 100th terms is 100. what is the difference between their 1 000th terms?
\nSolution:
\nHaving common difference ‘d’, the APs
\n1st set a, a+d, a+2d
\n2nd set b, b + d, b + 2d
\n100th<\/sup> term of 1st set – 100th<\/sup> term of 2nd set = 100
\n\u2234 a + 99d – (b + 99d) = 100
\na + 99d – b – 99d = 100
\na – b= 100
\nSimilarly.
\n1000th term of 1st set = 1 + 999d
\n1000 th term of 2nd set = b + 999d
\nTheir difference
\n= a + 999d – (b + 999d)
\n= a + 999d – b – 999d
\n= a – b.<\/p>\n

Arithmetic Progression 1.2 Exercise KSEEB Solutions Question 13.<\/strong>
\nHow many three-digit numbers are divisible by 7?
\nSolution:
\nFirst three-digit number divisible by 7
\n105 and the List number Is 994.
\n\u2234 AP is 105, 112. 119 994.
\na = 105, d = 112 – 105 = 7. an<\/sub> = 994.
\nn =?
\na + (n – 1)d = an<\/sub>
\n105 + (n – 1) 7 = 994
\n105 + 7n – 7 = 994
\n7n + 98 = 994
\n7n = 994 – 98
\n7n = 896
\n\\(n=\\frac{896}{7}\\)
\n\u2234 n = 128.
\n\u2234 Numbers with 3 digits divisible by 7 are 128.<\/p>\n

Arithmetic Progression Exercise 1.2 Solution KSEEB Solutions Question 14.<\/strong>
\nHow many multiples of 4 lie between 10 and 250?
\nSolution:
\nMultiples of 4. after 10 are 12, 16, 20….
\nMultiples of 4 upto 250 is 248
\n\u2234 A.P. is 12, 16, 20, …….. 248
\na = 12, d = 16 – 12 = 4
\nn =?
\na = a + (n – 1) d = 248
\n12 + 4n – 4 = 248
\n4n + 8 = 248
\n4n = 248 – 8
\n4n = 240
\n\\(n=\\frac{240}{4}\\)
\n\u2234 n = 60
\n\u2234 Multiples of 4 lie between 10 and 250 is 60.<\/p>\n

Exercise 1.2 Arithmetic Progression KSEEB Solutions Question 15.<\/strong>
\nFor what value of n\u2019. are the nth<\/sup> terms of two APs: 63, 65, 67, ……. and 3, 10, 17, ……….. equal?
\nSolution;
\n63, 65, 67,……….
\na = 63. d = 65 – 63 =2. an<\/sub> =?
\nnth<\/sup> term of this is
\nan<\/sub> = a + (n – 1)d
\n= 63 + (n – 1) 2
\n= 63 + 2n – 2
\nan<\/sub>= 2n + 61 …………….. (i)
\n3, 10, 17, ………….
\na = 3, d = 10 – 3 = 7, an<\/sub> =?
\nan<\/sub> = a + (n – 1)d
\n= 3 + (n – 1) 7
\n= 3 + 7n – 7
\nan<\/sub> = 7n \u2014 4 ………….(ii)
\nHere. nth<\/sup> terms of second AP are equal.
\n\u2234 equation (i) = equation (ii)
\n2n + 61 = 7n – 4
\n2n – 7n = -4 – 61
\n5n = 65
\n5n =65
\n\\(n=\\frac{65}{5}\\)
\n\u2234 n = 13
\n\u223413th terms of the two given APs are equal.<\/p>\n

Arithmetic Progression Ex 1.2 KSEEB Solutions Question 16.<\/strong>
\nDetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
\nSolution:
\na = 16, a7<\/sub> = a5<\/sub> + 12, A.P =?
\na7<\/sub> = a5<\/sub> + 12
\na + 6d = a + 4d + 12
\na + 6d – a – 4d = 12
\n2d = 12 \u2234 d = 6.
\na + 2d = 16
\na + 2(6) = 16
\na + 12 = 16
\n\u2234 a = 16 – 12 \u2234 a = 4
\na = 4, d = 6.
\n\u2234 A.P.:
\na, a + d, a + 2d, …………………
\n4, 4 + 6, 4 + 12, …………….
\n4, 10, 16, ………….<\/p>\n

Arithmetic Progression 1.2 KSEEB Solutions Question 17<\/strong>
\nFind the 20th term from the Last term of theAP: 3, 8, 13, …………., 253.
\nSolution:
\n3, 8, 13, ………….., 253
\na = 3. d = 8 – 3 = 5, an<\/sub> = 253
\n20th term from the last term of the AP starting from 253 =?
\n253, 258, 263, ………… a20<\/sub> =?
\na = 253, d = 258 – 253 = 5, n = 20
\nan<\/sub> = a + (n – 1) d
\na20<\/sub>= 253 + (20 – 1) 5
\n= 253 + 19 \u00d7 5
\n= 253 + 95
\n\u2234 a20<\/sub> = 348
\n\u2234 20th<\/sup> term from the last term of the AP is 348.<\/p>\n

Ex 1.2 Class 10 Arithmetic Progression KSEEB Solutions Question 18.<\/strong>
\nThe sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. FInd the first three terms of the AP.
\nSolution:
\na4<\/sub> + a8<\/sub> = 24 …………. (1)
\na6<\/sub> + a10<\/sub> = 44 ………… (2)
\nBut A.P is a, a + d, a + 2d
\nfrom equation (1).
\na4<\/sub> + a8<\/sub> = 24
\na + 3d + a + 7d = 24
\n2a + 10d = 24 ………….. (3)
\nfrom equation (2).
\na6<\/sub> + a10<\/sub> = 44
\na + 5d + a + 9d = 44
\n2a + 14d = 44 ………… (4)
\nSubtracting eqn. (4) from equation (3)
\n\"KSEEB
\n4d = 20
\n\\(d=20 \/ 4\\)
\n\u2234 d = 5
\nSubstituting the value of d in equation (3)
\n2a + 10d = 24
\n2a + 10(5) = 24
\n2a + 50 = 24
\n2a = 24 – 50
\n2a = -26
\na = 26\/2 ,
\n\u2234 a = -13 .
\n\u2234 AP: a, a + d, a + 2d, ………..
\n-13, -13 + 5, -13 + 2(5), ………
\n-14, -8, -3, …………<\/p>\n

10th Standard Arithmetic Progression Exercise 1.2 Question 19.<\/strong>
\nSubba Rao started work in 1995 at an annual salary of Rs. 5000 and received an Increment of Rs. 200 each year. in which year did his income reach Rs. 7000?
\nSolution:
\nPayment of Subba Rao in the year 1995 = Rs. 5000
\nincrement = Rs. 200
\n\u2234 The payment he received in the year 1996 is Rs. 5.200
\n\"KSEEB
\na = 5000, d = 5200 – 5000 = 200,
\nan<\/sub> = 7000, n=?
\na + (n – 1)d = an<\/sub>
\n5000 + (n – 1) 200 = 7000
\n5000 + 200n – 200 = 7000
\n200n + 4800 = 7000
\n200n = 7000 – 4800
\n200n = 2200
\n\\(n=\\frac{2200}{200}\\)
\n\u2234 n = 11
\nValue of \u2018n\u2019 is 11
\n\u2234 From 1995 to 10 years means 2005. his salary becomes Rs. 7,000.<\/p>\n

Ap Exercise 1.2 KSEEB Solutions Question 20.<\/strong>
\nRamkall saved Rs. 5 In the first week of a year and then Increased her weekly savings by Rs. 1.75. If in the n\u00fcI week. her weekly savings become Rs. 20.75. find \u2018n\u2019.
\nSolution:
\nSavings in the First week = Rs. 5.
\nSavings in the Second week 5 + 1.75 = Rs. 6.75
\nSavings in ‘nth<\/sup> week is Rs. 20.75
\n\u2234 A.P. 5, 6, 75 , …………. , 20.75
\na = 5, d = 6.75 – 5 = 1.75, an<\/sub> = 20.75.
\nn =?
\na + (n – 1) d = an<\/sub>
\n5 + (n – 1) (1.75) = 20.75
\n5 + 1.75n – 1.75 = 20.75
\n1 .75n + 3.25 = 20.75
\n1.75n = 20.75 – 3.25
\n1.75n = 17.5
\n\\(n=\\frac{17.5}{1.75}\\)
\n\\(n=\\frac{1750}{175}\\)
\n\u2234 n = 10
\n\u2234 In the 10th week, her savings becomes Rs. 20.75.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2, 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 1.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2. Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2 Arithmetic Progression Class 10 Exercise 1.2 …<\/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\/152"}],"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=152"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/152\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=152"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=152"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=152"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}