{"id":27519,"date":"2022-02-14T16:34:46","date_gmt":"2022-02-14T11:04:46","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=27519"},"modified":"2022-02-15T15:27:53","modified_gmt":"2022-02-15T09:57:53","slug":"ap-board-10th-class-maths-solutions-chapter-6-exercise-6-5-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-10th-class-maths-solutions-chapter-6-exercise-6-5-in-telugu\/","title":{"rendered":"AP Board 10th Class Maths Solutions Chapter 6 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.5"},"content":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions<\/a> Chapter 6 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.5 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 10th Class Maths Solutions 6th Lesson \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.5<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c2a\u0c4d\u0c30\u0c24\u0c3f \u0c17\u0c41\u0c23\u0c3f\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c15\u0c3f \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f\u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f\u0c28\u0c3f, n\u0c35 \u0c2a\u0c26\u0c2e\u0c41\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n(i) 3, \\(\\frac{3}{2}\\), \\(\\frac{3}{4}\\), \\(\\frac{3}{8}\\), ……..
\n\u0c38\u0c3e\u0c27\u0c28.
\n3, \\(\\frac{3}{2}\\), \\(\\frac{3}{4}\\), \\(\\frac{3}{8}\\), ……..
\n\u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = \\(\\frac{a_{2}}{a_{1}}=\\frac{\\frac{3}{2}}{3}=\\frac{3}{2} \\times \\frac{1}{3}=\\frac{1}{2}\\)
\nn\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = a rn – 1<\/sup>
\n= 3 \u00d7 (\\(\\frac{1}{2}\\))n – 1<\/sup><\/p>\n

(ii) 2, – 6, 18, – 54
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = \\(\\frac{a_{2}}{a_{1}}=\\frac{-6}{2}\\) = – 3
\nn\u0c35 \u0c2a\u0c26\u0c02 , an<\/sub> = a rn – 1<\/sup>
\n= 2 \u00d7 (- 3)n – 1.<\/sup><\/p>\n

\"AP<\/p>\n

(iii) – 1, – 3, – 9, – 27, ………..
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = \\(\\frac{a_{2}}{a_{1}}=\\frac{-3}{-1}\\) = 3
\nn\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = a rn – 1<\/sup>
\n= (- 1) \u00d7 3n – 1<\/sup> = – 3n – 1<\/sup><\/p>\n

(iv) 5, 2, \\(\\frac{4}{5}\\), \\(\\frac{8}{25}\\), …………..
\n\u0c38\u0c3e\u0c27\u0c28.
\n5, 2, \\(\\frac{4}{5}\\), \\(\\frac{8}{25}\\), …………..
\n\u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = \\(\\frac{a_{2}}{a_{1}}=\\frac{2}{5}\\)
\n1 \u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = a rn – 1<\/sup> = 5 \u00d7 (\\(\\frac{2}{5}\\))n – 1<\/sup><\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n5, 25, 125, ….. \u0c05\u0c28\u0c47 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u0c2f\u0c4a\u0c15\u0c4d\u0c15 10\u0c35, 1\u0c35 \u0c2a\u0c26\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n5, 25, 125, …………….
\na = 5, r = \\(\\frac{a_{2}}{a_{1}}=\\frac{25}{5}\\) = 5
\n10 \u0c35 \u0c2a\u0c26\u0c02 a10<\/sub> = a . r9<\/sup> = 5 \u00d7 59<\/sup> = 510<\/sup>
\nn\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = a . rn – 1<\/sup> = 5 \u00d7 (5)n – 1<\/sup>
\n= 51 + n – 1<\/sup> = 5n<\/sup><\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c32\u0c4b \u0c2a\u0c47\u0c30\u0c4d\u0c15\u0c4a\u0c28\u0c4d\u0c28 \u0c2a\u0c26\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n(i) a1<\/sub> = 9; r = \\(\\frac{1}{3}\\) \u0c05\u0c2f\u0c3f\u0c28 a7<\/sub> = ?
\n\u0c38\u0c3e\u0c27\u0c28.
\na1<\/sub> = 9; r = \\(\\frac{1}{3}\\)
\nar7<\/sub> = ar6<\/sup> – 9 \u00d7 (\\(\\frac{1}{3}\\) )6<\/sup>
\n= 32<\/sup> \u00d7 \\(\\frac{1}{3^{6}}\\)
\n= \\(\\frac{1}{3^{4}}=\\frac{1}{81}\\)<\/p>\n

(ii) a1<\/sub> = – 12; r = \\(\\frac{1}{3}\\); \u0c05\u0c2f\u0c3f\u0c28 a6<\/sub> = ?
\n\u0c38\u0c3e\u0c27\u0c28.
\na1<\/sub> = 12; r = \\(\\frac{1}{3}\\)
\na6<\/sub> = ar5<\/sup> = – 12(\\(\\frac{1}{3}\\))5<\/sup>
\n= \\(\\frac{-12}{3^{5}}=\\frac{-4}{3^{4}}=\\frac{4}{81}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n(i) 2, 8, 32, …….. \u0c17\u0c41\u0c23 \u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b \u0c0e\u0c28\u0c4d\u0c28\u0c35 \u0c2a\u0c26\u0c2e\u0c41 512 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c22 2, 8, 32, ……….. 512
\na = 2; r = \\(\\frac{a_{2}}{a_{1}}=\\frac{8}{2}\\) = 4, an<\/sub> = 512
\nan<\/sub> = a . rn – 1<\/sup> = 512
\n\u21d2 2 \u00d7 (4)n – 1<\/sup> = 512<\/p>\n

\"AP<\/p>\n

\u21d2 2 \u00d7 (22<\/sup>)n – 1<\/sup> = 29<\/sup>
\n\u21d2 2 \u00d7 22(n – 1)<\/sup>= 29<\/sup>
\n\u21d2 22n – 1<\/sup> = 29<\/sup>
\n2n – 1 = 9
\n2n = 9 + 1 = 10 ,
\nn = \\(\\frac{10}{2}\\) = 5
\n2, 8, 32, ….. \u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 5\u0c35 \u0c2a\u0c26\u0c02 512 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

(ii) \u221a3, 3, 3\u221a3, …………….. \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b \u0c0e\u0c28\u0c4d\u0c28\u0c35 \u0c2a\u0c26\u0c2e\u0c41 729 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u221a3, 3, 3\u221a3, …….. 729
\na = 3; r = \\(\\frac{a_{2}}{a_{1}}=\\frac{3}{\\sqrt{3}}=\\frac{\\sqrt{3} \\times \\sqrt{3}}{\\sqrt{3}}\\) = \u221a3
\nan<\/sub> = 729<\/p>\n

\"AP<\/p>\n

an<\/sub> = a rn – 1<\/sup> = 729
\n\u21d2 3 \u00d7 (\u221a3)n – 1<\/sup>= 729
\n\u21d2 (\u221a3)n<\/sup> = 729
\n\u21d2 3n\/2<\/sup> = 36<\/sup>
\n\u21d2 \\(\\frac{n}{2}\\) = 6
\n\u21d2 n = 12
\n\u221a3, 3, 3\u221a3, …………….. \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 12\u0c35 \u0c2a\u0c26\u0c02 729 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

\"AP<\/p>\n

(iii) \\(\\frac{1}{3}\\), \\(\\frac{1}{9}\\), \\(\\frac{1}{27}\\), ………… \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b \u0c0e\u0c28\u0c4d\u0c28\u0c35 \u0c2a\u0c26\u0c2e\u0c41 2187 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \\(\\frac{1}{3}\\), \\(\\frac{1}{9}\\), \\(\\frac{1}{27}\\), …………, \\(\\frac{1}{2187}\\)
\na = \\(\\frac{1}{3}\\), r = \\(\\frac{\\frac{1}{9}}{\\frac{1}{3}}=\\frac{1}{9} \\times \\frac{3}{1}=\\frac{1}{3}\\),
\nan<\/sub> = \\(\\frac{1}{2187}\\)
\nan<\/sub> = a . rn – 1<\/sup> = \\(\\frac{1}{2187}\\)
\n\u21d2 \\(\\frac{1}{3} \\times\\left(\\frac{1}{3}\\right)^{\\mathrm{n}-1}=\\frac{1}{2187}\\)<\/p>\n

\u21d2 \\(\\left(\\frac{1}{3}\\right)^{n}=\\left(\\frac{1}{3}\\right)^{7}\\)
\nn = 7<\/p>\n

\\(\\frac{1}{3}\\), \\(\\frac{1}{9}\\), \\(\\frac{1}{27}\\), ………… \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 7\u0c35 \u0c2a\u0c26\u0c02 \\(\\frac{1}{2187}\\) \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\u0c12\u0c15 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u0c2f\u0c4a\u0c15\u0c4d\u0c15 8\u0c35 \u0c2a\u0c26\u0c2e\u0c41 192 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f 2 \u0c05\u0c2f\u0c3f\u0c28 12\u0c35 \u0c2a\u0c26\u0c2e\u0c41\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n1\u0c35 \u0c2a\u0c26\u0c4d\u0c26\u0c24\u0c3f :
\n\u0c12\u0c15 \u0c17\u0c41\u0c23 \u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 8\u0c35 \u0c2a\u0c26\u0c02 a8<\/sub> = ar7<\/sup> = 192 ………. (1)
\n\u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = 2 \u0c28\u0c41 (1) \u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c17\u0c3e,
\na(2)7<\/sup> = 192
\na \u00d7 128 = 192
\n\u21d2 a = \\(\\frac{192}{128}=\\frac{3}{2}\\)
\n\u2234 12\u0c35 \u0c2a\u0c26\u0c02 a12<\/sub> = a r11<\/sup>
\n= \\(\\frac{3}{2}\\) \u00d7 (2)11<\/sup>
\n= 3 \u00d7 210<\/sup><\/p>\n

2\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\n\u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 8\u0c35 \u0c2a\u0c26\u0c02 a8<\/sub> = ar7<\/sup> = 192 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = 2 .
\n\u2234 12\u0c35 \u0c2a\u0c26\u0c02 a12<\/sub> = ar11<\/sup> = ar7<\/sup> \u00d7 r4<\/sup>
\n= 192 \u00d7 24<\/sup>
\n= 3 \u00d7 64 \u00d7 24<\/sup>
\n= 3 \u00d7 26<\/sup> \u00d7 24<\/sup>
\n= 3 \u00d7 210<\/sup><\/p>\n

3\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\n\u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 8\u0c35 \u0c2a\u0c26\u0c02 a8<\/sub> = ar7<\/sup> = 192 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = 2
\na9<\/sub> = 192 \u00d7 2 = 3 \u00d7 20 \u00d7 2 = 3 \u00d7 27
\na10<\/sub> = 3 \u00d7 27<\/sup> \u00d7 2 = 3 \u00d7 28<\/sup>
\na11<\/sub> = 3 \u00d7 28<\/sup> \u00d7 2 = 3 \u00d7 29<\/sup>
\na12<\/sub> = 3 \u00d7 29<\/sup> \u00d7 2 = 3 \u00d7 210<\/sup><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\u0c12\u0c15 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b \u0c28\u0c3e\u0c32\u0c4d\u0c17\u0c35 \u0c2a\u0c26\u0c2e\u0c41 \\(\\frac{2}{3}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 7\u0c35 \u0c2a\u0c26\u0c2e\u0c41 \\(\\frac{16}{81}\\) \u0c05\u0c2f\u0c3f\u0c28 \u0c06 \u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c17\u0c41\u0c23 \u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b \u0c28\u0c3e\u0c32\u0c4d\u0c17\u0c35 \u0c2a\u0c26\u0c2e\u0c41 a4<\/sub> = ar3<\/sup> = \\(\\frac{2}{3}\\) ………… (1)
\n7\u0c35 \u0c2a\u0c26\u0c2e\u0c41 a7<\/sub> = ar6<\/sup> = \\(\\frac{16}{81}\\) …………..(2)
\n(2) \u00f7 (1)
\n\u21d2 \\(\\frac{\\mathrm{ar}^{6}}{\\mathrm{ar}^{3}}=\\frac{\\frac{16}{81}}{\\frac{2}{3}}=\\frac{16}{81} \\times \\frac{3}{2}=\\frac{8}{27}\\) = \\(\\left(\\frac{2}{3}\\right)^{3}\\)<\/p>\n

\u21d2 r3<\/sup> = \\(\\left(\\frac{2}{3}\\right)^{3}\\)
\n\u2234 r = \\(\\frac{2}{3}\\) or
\nr = \\(\\frac{2}{3}\\) \u0c28\u0c3f (1) \u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c17\u0c3e,
\na\\(\\left(\\frac{2}{3}\\right)^{3}\\) = \\(\\frac{2}{3}\\)
\na \u00d7 \\(\\frac{8}{27}\\) = \\(\\frac{2}{3}\\)
\n\u21d2 a = \\(\\frac{2}{3} \\times \\frac{27}{8}=\\frac{9}{4}1\\)
\n\u2234 \u0c06 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c22 a, ar, ar2<\/sup>, ar3<\/sup>, ………….
\n\\(\\frac{9}{4}\\), \\(\\frac{9}{4} \\times \\frac{2}{3}\\), \\(\\frac{9}{4} \\times\\left(\\frac{2}{3}\\right)^{2}\\), \\(\\frac{9}{4} \\times\\left(\\frac{2}{3}\\right)^{3}\\), ………….
\n= \\(\\frac{9}{4}\\), \\(\\frac{3}{2}\\), 1, \\(\\frac{2}{3}\\), ………….<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n162, 54, 18, …… \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \\(\\frac{2}{81}\\), \\(\\frac{2}{27}\\), \\(\\frac{2}{9}\\) …… \u0c17\u0c41\u0c23 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32 1\u0c35 \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c2e\u0c41 \u0c05\u0c2f\u0c3f\u0c28 n \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 1\u0c35 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f 162, 54, 18, ……,
\na = 162, r = \\(\\frac{a_{2}}{a_{1}}=\\frac{54}{162}=\\frac{1}{3}\\)
\nn\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = a . rn – 1<\/sup>
\n= 162 . (\\(\\frac{1}{3}\\))n – 1<\/sup>
\n= \\(\\frac{162}{3^{n-1}}\\)
\n2\u0c35 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f
\n\\(\\frac{2}{81}\\), \\(\\frac{2}{27}\\), \\(\\frac{2}{9}\\), …………………..
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f\u0c2a\u0c26\u0c02 a = \\(\\frac{2}{81}\\), r = \\(\\frac{a_{2}}{a_{1}}\\)
\n= \\(\\frac{\\frac{2}{27}}{\\frac{2}{81}}=\\frac{2}{27} \\times \\frac{81}{2}\\) = 3<\/p>\n

n \u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = \\(\\frac{2}{81}\\) (3)n – 1<\/sup> = \\(\\frac{2 \\times 3^{n-1}}{81}\\)
\n\u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c17\u0c41\u0c23\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32 n \u0c35 \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c2e\u0c41.
\n\\(\\frac{162}{3^{n-1}}=\\frac{2 \\times 3^{n-1}}{81}\\)
\n2 \u00d7 3n – 1<\/sup> \u00d7 3n – 1<\/sup> = 162 \u00d7 81 (\u0c05\u0c21\u0c4d\u0c21\u0c17\u0c41\u0c23\u0c15\u0c3e\u0c30\u0c2e\u0c41 \u0c1a\u0c47\u0c2f\u0c17\u0c3e)
\n32n – 2<\/sup> = \\(\\frac{162 \\times 81}{2}\\) = 81 \u00d7 81
\n32n – 2<\/sup> = 34<\/sup> \u00d7 34<\/sup> = 38<\/sup>
\n32n – 2<\/sup> = 38<\/sup>
\n\u2234 2n – 2 = 8
\n2n = 8 + 2 = 10
\nn = \\(\\frac{10}{2}\\) = 5
\n\u2234 n = 5<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions Chapter 6 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.5 Textbook Exercise Questions and Answers. AP State Syllabus 10th Class Maths Solutions 6th Lesson \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.5 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c2a\u0c4d\u0c30\u0c24\u0c3f \u0c17\u0c41\u0c23\u0c3f\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c15\u0c3f \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f\u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f\u0c28\u0c3f, n\u0c35 \u0c2a\u0c26\u0c2e\u0c41\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41. (i) 3, , , , …….. \u0c38\u0c3e\u0c27\u0c28. 3, , , , …….. \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c28\u0c3f\u0c37\u0c4d\u0c2a\u0c24\u0c4d\u0c24\u0c3f r = …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/27519"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=27519"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/27519\/revisions"}],"predecessor-version":[{"id":27619,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/27519\/revisions\/27619"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=27519"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=27519"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=27519"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}