{"id":127329,"date":"2021-09-14T14:21:21","date_gmt":"2021-09-14T08:51:21","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=127329"},"modified":"2021-09-15T14:19:27","modified_gmt":"2021-09-15T08:49:27","slug":"rd-sharma-class-10-solutions-chapter-9-exercise-9-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.5 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127362\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.5.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-9-Exercise-9.5-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5:\u00a0<\/strong>The selection of terms in an A.P. is the main focus of the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Maths Solutions<\/strong><\/a> chapter 9 exercise 9.5. Kopykitab created answers to these exercise problems with the primary goal of dispelling doubts and enhancing knowledge in A.P. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-arithmetic-progressions\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 9<\/strong><\/a> Exercise 9.5 PDF is also available for download.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d89f3a04b9b\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d89f3a04b9b\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/#download-rd-sharma-class-10-solutions-chapter-9-exercise-95-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-9-exercise-95-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 9 Exercise 9.5- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 9 Exercise 9.5- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/#faqs-on-rd-sharma-class-10-solutions-chapter-9-exercise-95\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5\">FAQs on RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/#is-the-rd-sharma-class-10-solutions-chapter-9-exercise-95-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 available on the Kopykitab website?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/#what-are-the-key-benefits-of-learning-rd-sharma-solutions-for-class-10-maths-chapter-9-exercise-95\" title=\"What are the key benefits of learning RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.5?\">What are the key benefits of learning RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.5?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-9-exercise-95-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-9-exercise-95-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-9-Exercise-9.5.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-9-Exercise-9.5.pdf\">RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-9-exercise-95-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 9 Exercise 9.5- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Find the value of x for which (8x + 4), (6x \u2013 2) and (2x + 7) are in A.P.<br \/>Solution:<br \/>(8x + 4), (6x \u2013 2) and (2x + 7) are in A.P.<br \/>(6x \u2013 2) \u2013 (8x + 4) = (2x + 7) \u2013 (6x \u2013 2)<br \/>\u21d2 6x \u2013 2 \u2013 8x \u2013 4 = 2x + 7 \u2013 6x + 2<br \/>\u21d2 -2x \u2013 6 = -4x + 9<br \/>\u21d2 -2x + 4x = 9 + 6<br \/>\u21d2 2x = 15<br \/>Hence x =\u00a0152<\/p>\n<p>Question 2.<br \/>If x + 1, 3x and 4x + 2 are in A.P., find the value of x.<br \/>Solution:<br \/>x + 1, 3x and 4x + 2 are in A.P.<br \/>3x \u2013 x \u2013 1 = 4x + 2 \u2013 3x<br \/>\u21d2 2x \u2013 1 = x + 2<br \/>\u21d2 2x \u2013 x = 2 + 1<br \/>\u21d2 x = 3<br \/>Hence x = 3<\/p>\n<p>Question 3.<br \/>Show that (a \u2013 b)\u00b2, (a\u00b2 + b\u00b2) and (a + b)\u00b2 are in A.P.<br \/>Solution:<br \/>(a \u2013 b)\u00b2, (a\u00b2 + b\u00b2) and (a + b)\u00b2 are in A.P.<br \/>If 2 (a\u00b2 + b\u00b2) = (a \u2013 b)\u00b2 + (a + b)\u00b2<br \/>If 2 (a\u00b2 + b\u00b2) = a\u00b2 + b\u00b2 \u2013 2ab + a\u00b2 + b\u00b2 + 2ab<br \/>If 2 (a\u00b2 + b\u00b2) = 2a\u00b2 + 2b\u00b2 = 2 (a\u00b2 + b\u00b2)<br \/>Which is true<br \/>Hence proved.<\/p>\n<p>Question 4.<br \/>The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceed the second term by 6, find three terms.<br \/>Solution:<br \/>Let the three terms of an A.P. be a \u2013 d, a, a + d<br \/>Sum of three terms = 21<br \/>\u21d2 a \u2013 d + a + a + d = 21<br \/>\u21d2 3a = 21<br \/>\u21d2 a = 7<br \/>and product of the first and 3rd = 2nd term + 6<br \/>\u21d2 (a \u2013 d) (a + d) = a + 6<br \/>a\u00b2 \u2013 d\u00b2 = a + 6<br \/>\u21d2 (7 )\u00b2 \u2013 d\u00b2 = 7 + 6<br \/>\u21d2 49 \u2013 d\u00b2 = 13<br \/>\u21d2 d\u00b2 = 49 \u2013 13 = 36<br \/>\u21d2 d\u00b2 = (6)\u00b2<br \/>\u21d2 d = 6<br \/>Terms are 7 \u2013 6, 7, 7 + 6 \u21d2 1, 7, 13<\/p>\n<p>Question 5.<br \/>Three numbers are in A.P. If the sum of these numbers is 27 and the product 648, find the numbers.<br \/>Solution:<br \/>Let the three numbers of an A.P. be a \u2013 d, a, a + d<br \/>According to the conditions,<br \/>Sum of these numbers = 27<br \/>a \u2013 d + a + a + d = 27<br \/>\u21d2 3a = 27<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/41920188894_2e0c2fe832_o.png\" alt=\"RD Sharma Class 10 Chapter 9 Arithmetic Progressions \" width=\"327\" height=\"302\" \/><\/p>\n<p>Question 6.<br \/>Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.<br \/>Solution:<br \/>Let the four terms of an A.P. be (a \u2013 3d), (a \u2013 d), (a + d) and (a + 3d)<br \/>Now according to the condition,<br \/>Sum of these terms = 50<br \/>\u21d2 (a \u2013 3d) + (a \u2013 d) + (a + d) + (a + 3d) = 50<br \/>\u21d2 a \u2013 3d + a \u2013 d + a + d + a \u2013 3d= 50<br \/>\u21d2 4a = 50<br \/>\u21d2 a =\u00a0252<br \/>and greatest number = 4 x least number<br \/>\u21d2 a + 3d = 4 (a \u2013 3d)<br \/>\u21d2 a + 3d = 4a \u2013 12d<br \/>\u21d2 4a \u2013 a = 3d + 12d<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/27769852227_30587918bc_o.png\" alt=\"Arithmetic Progressions Class 10 RD Sharma \" width=\"352\" height=\"511\" \/><\/p>\n<p>Question 7.<br \/>The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1756\/27769852747_b8dc961490_o.png\" alt=\"RD Sharma Class 10 Solutions Arithmetic Progressions \" width=\"355\" height=\"383\" \/><\/p>\n<p>Question 8.<br \/>Divide 56 into four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5: 6. [CBSE 2016]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1732\/27769853717_9f28159742_o.png\" alt=\"RD Sharma Class 10 Solutions Arithmetic Progressions Exercise 9.5 \" width=\"331\" height=\"371\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1756\/27769853057_e85eda7117_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions \" width=\"238\" height=\"303\" \/><\/p>\n<p>Question 9.<br \/>The angles of a quadrilateral are in A.P. whose common difference is 10\u00b0. Find the angles.<br \/>Solution:<br \/>Let the four angles of a quadrilateral which are in A.P., be<br \/>a \u2013 3d, a \u2013 d, a + d, a + 3d<br \/>Common difference = 10\u00b0<br \/>Now sum of angles of a quadrilateral = 360\u00b0<br \/>a \u2013 3d + a \u2013 d + a + d + a + 3d = 360\u00b0<br \/>\u21d2 4a = 360\u00b0<br \/>\u21d2 a = 90\u00b0<br \/>and common difference = (a \u2013 d) \u2013 (a \u2013 3d) = a \u2013 d \u2013 a + 3d = 2d<br \/>2d = 10\u00b0<br \/>\u21d2 d = 5\u00b0<br \/>Angles will be<br \/>a \u2013 3d = 90\u00b0 \u2013 3 x 5\u00b0 = 90\u00b0 \u2013 15\u00b0 = 75\u00b0<br \/>a \u2013 d= 90\u00b0 \u2013 5\u00b0 = 85\u00b0<br \/>a + d = 90\u00b0 + 5\u00b0 = 95\u00b0<br \/>and a + 3d = 90\u00b0 + 3 x 5\u00b0 = 90\u00b0 + 15\u00b0= 105\u00b0<br \/>Hence the angles of the quadrilateral will be<br \/>75\u00b0, 85\u00b0, 95\u00b0 and 105\u00b0<\/p>\n<p>Question 10.<br \/>Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623. [NCERT Exemplar]<br \/>Solution:<br \/>Let the three parts of the number 207 are (a \u2013 d), a and (a + d), which are in A.P.<br \/>Now, by given condition,<br \/>\u21d2 Sum of these parts = 207<br \/>\u21d2 a \u2013 d + a + a + d = 207<br \/>\u21d2 3a = 207<br \/>a = 69<br \/>Given that, product of the two smaller parts = 4623<br \/>\u21d2 a (a \u2013 d) = 4623<br \/>\u21d2 69 (69 \u2013 d) = 4623<br \/>\u21d2 69 \u2013 d = 67<br \/>\u21d2 d = 69 \u2013 67 = 2<br \/>So, first part = a \u2013 d = 69 \u2013 2 = 67,<br \/>Second part = a = 69<br \/>and third part = a + d = 69 + 2 = 71<br \/>Hence, required three parts are 67, 69, 71.<\/p>\n<p>Question 11.<br \/>The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles. [NCERT Exemplar]<br \/>Solution:<br \/>Given that, the angles of a triangle are in A.P.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/27769854237_69314a64c9_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 9 Arithmetic Progressions \" width=\"352\" height=\"489\" \/><\/p>\n<p>Question 12.<br \/>The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7: 15. Find the number. [NCERT Exemplar]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1740\/27769854847_c6cc6fc824_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 9 Arithmetic Progressions \" width=\"340\" height=\"360\" \/><br \/>or, d = \u00b1 2<br \/>So, when a = 8, d = 2,<br \/>the numbers are 2, 6, 10, 14.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-9-exercise-95\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631607750215\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-9-exercise-95-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631608278482\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-key-benefits-of-learning-rd-sharma-solutions-for-class-10-maths-chapter-9-exercise-95\"><\/span>What are the key benefits of learning RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.5?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 answers were quite helpful. This makes it simple to clear any questions about arithmetic progression. Easily answer all of the questions that the Class 10 students have given for exercise.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631608449408\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-9-exercise-95-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5:\u00a0The selection of terms in an A.P. is the main focus of the RD Sharma Class 10 Maths Solutions chapter 9 exercise 9.5. Kopykitab created answers to these exercise problems with the primary goal of dispelling doubts and enhancing knowledge in A.P. The RD Sharma Solutions for &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.5 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-9-exercise-9-5\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.5 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":127362,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127329"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=127329"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127329\/revisions"}],"predecessor-version":[{"id":127950,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127329\/revisions\/127950"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127362"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=127329"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=127329"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=127329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}