{"id":27354,"date":"2022-02-11T19:00:43","date_gmt":"2022-02-11T13:30:43","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=27354"},"modified":"2022-02-11T19:35:41","modified_gmt":"2022-02-11T14:05:41","slug":"ap-board-10th-class-maths-solutions-chapter-6-exercise-6-2-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-10th-class-maths-solutions-chapter-6-exercise-6-2-in-telugu\/","title":{"rendered":"AP Board 10th Class Maths Solutions Chapter 6 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.2"},"content":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions<\/a> Chapter 6 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.2 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 6.2<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c2e\u0c41 a, \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c2d\u0c47\u0c26\u0c2e\u0c41 d, n\u0c35 \u0c2a\u0c26\u0c2e\u0c41 a, \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c4d\u0c1f\u0c3f\u0c15\u0c28\u0c41 \u0c2a\u0c42\u0c30\u0c3f\u0c02\u0c2a\u0c41\u0c2e\u0c41. – AS,, AS,,<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

\u0c38\u0c3e\u0c27\u0c28.<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n(i) 10, 7, 4, …… \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 30\u0c35 \u0c2a\u0c26\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P. = 10, 7, 4, …………….
\na1<\/sub> = 10;
\nd = a2<\/sub> – a1<\/sub> = 7 – 10 = – 3,
\nn= 30
\nan<\/sub> = a + (n – 1)
\na30<\/sub> = 10 + (30 – 1) (- 3)
\n= 10 + 29 (- 3)
\n= 10 – 87 = – 77
\n\u2234 a30<\/sub> = – 77.<\/p>\n

(ii) – 3, \\(-\\frac{1}{2}\\), – 2, ………….. \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 11\u0c35 \u0c2a\u0c26\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P. = – 3, \\(-\\frac{1}{2}\\), – 2, …………..
\na = – 3; d = a2<\/sub> – a1<\/sub> = 3 – 3)
\n= \\(-\\frac{1}{2}\\) + 3 = 2\\(\\frac{1}{2}\\) = \\(\\frac{5}{2}\\)
\nn = 11
\n\u2234 an<\/sub> = a + (n – 1) d
\na11<\/sub> = – 3 + (11 – 1) (\\(\\frac{5}{2}\\))
\n= – 3 + 10(\\(\\frac{5}{2}\\))
\n= – 3 + 25 = 22
\n\u2234 a11<\/sub>\u00a0= 22.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n(i) a1<\/sub> = 2; a3<\/sub> = 26, \u0c05\u0c2f\u0c3f\u0c28 a2<\/sub> \u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\na1<\/sub> = a = 2
\na3<\/sub> = 26
\nan<\/sub> = a + (n – 1) d
\na3<\/sub> = 2 + (3 – 1) 4
\n= 2d= 26 – 2 = 24
\nd = \\(\\frac{24}{2}\\) = 12
\n\u2234 a2<\/sub> = a + d = 2 + 12 = 14.<\/p>\n

\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\na1<\/sub>, a2<\/sub>, a3<\/sub> \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41 \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 a1<\/sub> = 2, a3<\/sub> = 26
\n\u2234 2, a2<\/sub>, 26 \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41.
\na2<\/sub> – 2 = 26 – a2<\/sub>
\n\u2234 a2<\/sub> + a2<\/sub> = 26 + 2
\n2a2<\/sub> = 28
\n\u2234 a2<\/sub> = \\(\\frac{28}{2}\\) = 14.<\/p>\n

\u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\na, b, c \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c09\u0c02\u0c1f\u0c47 b = \\(\\frac{a+c}{2}\\)
\n2, a2<\/sub>, 26 \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41.
\n\u2234 a2<\/sub> = \\(\\frac{2+26}{2}\\) = \\(\\frac{28}{2}\\) = 14.<\/p>\n

\"AP<\/p>\n

(ii) a2<\/sub> = 13; a4<\/sub> = 3 \u0c05\u0c2f\u0c3f\u0c28 a1<\/sub>, a3<\/sub> \u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\na2<\/sub> = a + d = 13 …….. (1)
\na4<\/sub> = a + 3d = 3 …….. (2)
\n(1), (2) \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41\u0c32\u0c41 \u0c38\u0c3e\u0c27\u0c3f\u0c02\u0c1a\u0c17\u0c3e,<\/p>\n

\"AP<\/p>\n

\u21d2 d = \\(-\\frac{10}{2}\\) = – 5
\na1<\/sub> = a2<\/sub> – d = 13 – (-5) = 13 + 5 = 18
\na3<\/sub> = a2<\/sub> + d = 13 + (- 5) = 8
\n\u2234 a1<\/sub> = 18 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 a3<\/sub> = 8.<\/p>\n

\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\na1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub> \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41 \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c2e\u0c41 a2<\/sub> = 13, a4<\/sub> = 3
\n\u2234 a1<\/sub>, 13, a3<\/sub>, 3 \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41
\n\u2234 13 – a1<\/sub> = a3<\/sub> – 13 ……. (1) \u0c2e\u0c30\u0c3f\u0c2f\u0c41
\na3<\/sub> – 13 = 3 – a3<\/sub> ……….. (2)
\n(2) \u21d2 2a3<\/sub> = 16
\na3<\/sub> = \\(\\frac{16}{2}\\) = 8
\na3<\/sub> = 8 \u0c28\u0c3f (1) \u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c17\u0c3e,
\n13 – a1<\/sub> = 8 – 13
\n– a1<\/sub> = – 5 – 13 = – 18
\n\u2234 a1<\/sub> = 18
\n\u2234 a1<\/sub> = 18 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 a3<\/sub> = 8.<\/p>\n

\u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\na1<\/sub>, 13, a3<\/sub>, 3 \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c15\u0c32\u0c35\u0c41.
\n\u2234 13, a3<\/sub>, 3 \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c2e\u0c42\u0c21\u0c41 \u0c35\u0c30\u0c41\u0c38 \u0c2a\u0c26\u0c3e\u0c32\u0c41.
\n\u2234 a3<\/sub> = \\(\\frac{13+3}{2}=\\frac{16}{2}\\) = 8
\n[a, b, c \u0c32\u0c41 A.P. \u0c32\u0c4b \u0c09\u0c02\u0c1f\u0c47 b = \\(\\frac{a+c}{2}\\))
\n\u2234 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c2d\u0c47\u0c26\u0c02 d = a3<\/sub> – a2<\/sub> = 8 – 13 = – 5
\n\u2234 a1<\/sub> = a2<\/sub> – d = 13 – (- 5) = 13 + 5 = 18.
\n\u2234 a1<\/sub> = 18 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 a3<\/sub> = 8.<\/p>\n

\"AP<\/p>\n

(iii) a1<\/sub> = 5, a4<\/sub> = 91\/2 \u0c05\u0c2f\u0c3f\u0c28 a2<\/sub>, a3<\/sub> \u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\na1<\/sub> = a = 5.
\nan<\/sub> = a + (n – 1) d
\na4<\/sub> = 5 + 3d = 9\\(\\frac{1}{2}\\) = \\(\\frac{19}{2}\\)
\n3d = \\(\\frac{19}{2}\\) – 5 = \\(\\frac{19-10}{2}\\) = \\(\\frac{9}{2}\\)
\n\u2234 d = \\(\\frac{9}{2} \\times \\frac{1}{3}=\\frac{3}{2}\\)
\n\u2234 a2<\/sub> = a + d
\n= 5 + \\(\\frac{3}{2}\\) = \\(\\frac{13}{2}\\)
\na3<\/sub> = a2<\/sub> + d
\n= \\(\\frac{13}{2}\\) + \\(\\frac{3}{2}\\) = \\(\\frac{16}{2}\\) = 8<\/p>\n

(iv) a1<\/sub> = – 4; a6<\/sub> = 6, \u0c05\u0c2f\u0c3f\u0c28 a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub> \u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41. .
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\na1<\/sub> = a = – 4
\na6<\/sub> = a + 5d = 6
\n(- 4) + 5d = 6
\n\u21d2 5d = 6 + 4 = 10
\n\u21d2 d = \\(\\frac{10}{5}\\) = 2
\n\u2234 a2<\/sub> = – 4 + 2 = – 2
\na3<\/sub> = – 2 + 2 = 0
\na4<\/sub> = 0 + 2 = 2
\na5<\/sub> = 2 + 2 = 4<\/p>\n

\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c26\u0c24\u0c3f :
\n\u0c12\u0c15 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b n\u0c35 \u0c2a\u0c26\u0c02 an<\/sub>, m\u0c35 \u0c2a\u0c26\u0c02 am<\/sub> \u0c05\u0c2f\u0c3f\u0c28 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f\u0c2d\u0c47\u0c26\u0c02
\nd = \\(\\frac{a_{m}-a_{n}}{m-n}\\)
\na1<\/sub> = – 4, a6<\/sub> = 6, n = 1; m = 6
\nd = \\(\\frac{6-(-4)}{6-1}=\\frac{10}{5}\\) = 2
\n\u2234 a2<\/sub> = a1<\/sub> + d = – 4 + 2 = – 2
\na3<\/sub> = – 2 + 2 = 0
\na4<\/sub> = 0 + 2 = 2
\na5<\/sub> = 2 + 2 = 4.<\/p>\n

\"AP<\/p>\n

(v) a2<\/sub> = 38; a6<\/sub> = – 22, \u0c05\u0c2f\u0c3f\u0c28 a1<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub> \u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\na2<\/sub> = a + d = 38 ……………. (1)
\na6<\/sub> = a + 5d = -22 …………… (2)
\n(2) – (1)<\/p>\n

\"AP<\/p>\n

\u2234 a1<\/sub> = a2<\/sub> – d = 38 – (- 15) = 38 + 15 = 53
\na3<\/sub> = a2<\/sub> + 4 = 38 + (-15) = 23
\na4<\/sub> = 23 + (- 15) = 8
\na5<\/sub> = 8 + (- 15) = – 7.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n3, 8, 13, 18, … \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b \u0c0e\u0c28\u0c4d\u0c28\u0c35 \u0c2a\u0c26\u0c2e\u0c41 78 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f : 3, 8, 13, 18, ……. 78 –
\na = 3; d = a2<\/sub> – a1<\/sub> = 8 – 3 = 5,
\nan<\/sub> = 78
\nan<\/sub> = a + (n – 1) 4 = 78 .
\n\u21d2 3 + (n – 1) (5) = 78
\n\u21d2 3 + 5n – 5 = 78
\n\u21d2 5n – 2 = 78
\n\u21d2 5n = 78 + 2 = 80
\n\u21d2 n = \\(\\frac{80}{5}\\) = 16
\n\u2234 16 \u0c35 \u0c2a\u0c26\u0c2e\u0c41 78 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c32\u0c4b\u0c28\u0c3f \u0c2a\u0c26\u0c3e\u0c32 \u0c38\u0c02\u0c16\u0c4d\u0c2f\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n(i) 7, 13, 19, . . . , 205
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c4d\u0c26\u0c24\u0c3f :
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P : 7, 13, 19, …………, 205
\na = 7; d = a2<\/sub> – a1<\/sub> = 13 – 7 = 6,
\nan<\/sub> = 205
\nan<\/sub> = a + (n – 1) d = 205
\n7 + (n – 1) 6 = 205
\n7 + 6n – 6 = 205
\n6n + 1 = 205
\n6n = 205 – 1 = 204
\n\u21d2 n = \\(\\frac{204}{6}\\) = 34
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P \u0c32\u0c4b 34 \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c09\u0c02\u0c1f\u0c3e\u0c2f\u0c3f.<\/p>\n

\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f:
\nd = \\(\\frac{a_{m}-a_{n}}{m-n}=\\frac{a_{n}-a_{m}}{n-m}\\)
\na1<\/sub> = 7, an<\/sub> = 1, am<\/sub> = 205 \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\nd = 13 – 7 = 6
\n6 = \\(\\frac{205-7}{n-1}\\)
\n\u21d2 n – 1 = \\(\\frac{198}{6}\\) = 33
\n\u2234 n = 33 + 1 = 34.<\/p>\n

\"AP<\/p>\n

(ii) 18, 15\\(\\frac{1}{2}\\), 13, ………, – 47
\n\u0c38\u0c3e\u0c27\u0c28. \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f : –
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P: 18, 15\\(\\frac{1}{2}\\), 1.3 ………….. – 47
\na = 18, d = a2<\/sub> – a1<\/sub>
\n= 15\\(\\frac{1}{2}\\) – 18
\n= – 2\\(\\frac{1}{2}\\) = – \\(\\frac{5}{2}\\)
\nan<\/sub> = – 47
\nan<\/sub> = a + (n – 1) d = – 47
\n= 18 + (n – 1) \u00d7 (- \\(\\frac{5}{2}\\)) = – 47
\n(n – 1) (- \\(\\frac{5}{2}\\)) = – 47 – 18 = – 65
\n\\(\\frac{-5 n+5}{2}\\) = – 65
\n– 5n + 5 = – 130
\n– 5n = – 130 – 5 = – 135
\n5n = 135
\n\u21d2 n = \\(\\frac{135}{5}\\) = 27
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P \u0c32\u0c4b 27 \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c09\u0c02\u0c1f\u0c3e\u0c2f\u0c3f.<\/p>\n

\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\nam<\/sub> = 18, d = 35; an<\/sub> = – 47
\nd = \\(\\frac{a_{n}-a_{m}}{n-m}\\)<\/p>\n

\u21d2 \\(\\frac{-5}{2}=\\frac{-47-18}{n-1}\\)<\/p>\n

\u21d2 \\(\\frac{-5}{2}=\\frac{-65}{n-1}\\)<\/p>\n

\u21d2 \\(\\frac{5}{2}=\\frac{65}{n-1}\\)
\nn – 1 = 65 \u00d7 \\(\\frac{2}{5}\\)
\n\u2234 n = 26 + 1 = 27.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n11, 8, 5, 2… \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b ‘- 150’ \u0c12\u0c15 \u0c2a\u0c26\u0c02\u0c17\u0c3e \u0c09\u0c02\u0c1f\u0c41\u0c02\u0c26\u0c4b \u0c32\u0c47\u0c26\u0c4b \u0c2a\u0c30\u0c3f\u0c36\u0c40\u0c32\u0c3f\u0c02\u0c1a\u0c41\u0c2e\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f 11, 8, 5, 2, …… \u0c32\u0c4b n \u0c35 \u0c2a\u0c26\u0c02 – 150 \u0c05\u0c28\u0c41\u0c15\u0c41\u0c02\u0c26\u0c3e\u0c2e\u0c41.
\n\u0c05\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, a = 11, d = a2<\/sub> – a1<\/sub> = 8 – 11 = – 3 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 an<\/sub> = – 150
\nan<\/sub> = a + (n – 1) d = – 150
\n\u21d2 11 + (n – 1)X (- 3) = – 150
\n\u21d2 11 – 3n + 3 = – 150
\n\u21d2 – 3n = – 150 – 14
\n\u21d2 – 3n = – 164
\n\u21d2 3n = 164
\n\u21d2 n = \\(\\frac{164}{3}\\) ………… (2)
\n\u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b\u0c28\u0c3f \u0c2a\u0c26\u0c3e\u0c32 \u0c38\u0c02\u0c16\u0c4d\u0c2f n \u0c0e\u0c32\u0c4d\u0c32\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c42 \u0c12\u0c15 \u0c38\u0c39\u0c1c \u0c38\u0c02\u0c16\u0c4d\u0c2f.
\n\u0c15\u0c3e\u0c28\u0c3f n = \\(\\frac{164}{3}\\) \u0c38\u0c39\u0c1c\u0c38\u0c02\u0c16\u0c4d\u0c2f \u0c15\u0c3e\u0c26\u0c41.
\n\u0c15\u0c3e\u0c35\u0c41\u0c28 11, 8, 5, 2, ……. \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b – 150 \u0c12\u0c15 \u0c2a\u0c26\u0c02\u0c17\u0c3e \u0c09\u0c02\u0c21\u0c26\u0c41.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n\u0c12\u0c15 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 11\u0c35 \u0c2a\u0c26\u0c2e\u0c41 38 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 16\u0c35 \u0c2a\u0c26\u0c2e\u0c41 78 \u0c05\u0c2f\u0c3f\u0c28 31\u0c35 \u0c2a\u0c26\u0c2e\u0c41\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\na11<\/sub> = 38 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 a16<\/sub> = 73, a31<\/sub> = ?
\n::. an<\/sub> = a + (n – 1) d
\na11<\/sub> = a + 10d = 38 …………. (1)
\na16 <\/sub> = a + 15d = 73 ,
\n(2) – (1)<\/p>\n

\"AP<\/p>\n

d = 7 \u0c28\u0c41 (1) \u0c32\u0c4b \u0c30\u0c3e \u0c2f\u0c17\u0c3e,
\na + 70 = 38
\n\u21d2 a = 38 – 70 = – 32
\n31\u0c35 \u0c2a\u0c26\u0c02 a31<\/sub> = a + 30d
\n= – 32 + 30 (7)
\n= – 32 + 210 = 178
\n\u2234 31\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = 178.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n\u0c12\u0c15 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c3f\u0c32\u0c4b 3\u0c35, 9\u0c35 \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c35\u0c30\u0c41\u0c38\u0c17\u0c3e 4, – 8 \u0c05\u0c2f\u0c3f\u0c28 \u0c0e\u0c28\u0c4d\u0c28\u0c35 \u0c2a\u0c26\u0c2e\u0c41 ” (\u0c38\u0c41\u0c28\u0c4d\u0c28) \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c12\u0c15 A.P \u0c32\u0c4b 3\u0c35 \u0c2a\u0c26\u0c02
\na3<\/sub> = a + 2d = 4 ……….(1)
\n9\u0c35 \u0c2a\u0c26\u0c02 a9<\/sub> = a + 8d = – 8 …………(2)
\n(2) – (1) \u21d2<\/p>\n

\"AP<\/p>\n

\u21d2 d = \\(\\frac{12}{6}\\) = – 2
\n\u21d2 d = – 2 …………. (3)
\n\u2234 4\u0c35 \u0c2a\u0c26\u0c02 a4<\/sub> = a3<\/sub> + d = 4 + (- 2) = 2
\na5<\/sub> = a4<\/sub> + d = 2 + (- 2) = 0
\n\u2234 5\u0c35 \u0c2a\u0c26\u0c02 \u0c38\u0c41\u0c28\u0c4d\u0c28 (0) \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

(\u0c32\u0c47\u0c26\u0c3e)<\/p>\n

(3) \u21d2 d = – 2 \u0c28\u0c41 (1) \u0c32\u0c4b \u0c30\u0c3e\u0c2f\u0c17\u0c3e,
\na + 2(- 2) = 4
\n\u21d2 a – 4 = 4
\n\u21d2 a = 8
\nan<\/sub> = 0 \u0c05\u0c2f\u0c4d\u0c2f\u0c47\u0c1f\u0c1f\u0c4d\u0c32\u0c41 n \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c3e\u0c32\u0c3f.
\nan<\/sub> = a + (n – 1) d = 0
\n8 + (n – 1) (- 2) = 0
\n8 – 2n + 2 = 0
\n10 = 2n
\n\u21d2 \\(\\frac{10}{2}\\) = 5
\n\u2234 n = 5
\n\u0c15\u0c3e\u0c35\u0c41\u0c28 5\u0c35 \u0c2a\u0c26\u0c02 ‘0’ (\u0c38\u0c41\u0c28\u0c4d\u0c28) \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 9.
\n\u0c12\u0c15 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 17\u0c35 \u0c2a\u0c26\u0c2e\u0c41 10\u0c35 \u0c2a\u0c26\u0c02 \u0c15\u0c02\u0c1f\u0c47 7 \u0c0e\u0c15\u0c4d\u0c15\u0c41\u0c35. \u0c05\u0c2f\u0c3f\u0c28 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c2d\u0c47\u0c26\u0c02 \u0c0e\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c12\u0c15 A.P \u0c32\u0c4b 17\u0c35 \u0c2a\u0c26\u0c02 a17<\/sub> = a + 16d
\n10\u0c35 \u0c2a\u0c26\u0c02 a10<\/sub> = a + 9d
\n\u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02, a17<\/sub> = a10<\/sub> + 7
\na + 16d = (a + 9d) + 7
\na + 16d – a – 9d = 7
\n7d = 7
\n\u21d2 d = \\(\\frac{7}{7}\\) = 1
\n\u2234 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f\u0c2d\u0c47\u0c26\u0c02 d = 1. .<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 10.
\n\u0c30\u0c46\u0c02\u0c21\u0c41 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c2d\u0c47\u0c26\u0c02 \u0c38\u0c2e\u0c3e\u0c28\u0c2e\u0c41. \u0c35\u0c3e\u0c28\u0c3f 100\u0c35 \u0c2a\u0c26\u0c3e\u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c2d\u0c47\u0c26\u0c02 100 \u0c05\u0c2f\u0c3f\u0c28 \u0c35\u0c3e\u0c28\u0c3f 1000\u0c35 \u0c2a\u0c26\u0c3e\u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c2d\u0c47\u0c26\u0c2e\u0c46\u0c02\u0c24 ?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = a
\n\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = b
\n\u0c30\u0c46\u0c02\u0c21\u0c41 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f\u0c2d\u0c47\u0c26\u0c02 = d \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c3f 100\u0c35 \u0c2a\u0c26\u0c02 a100<\/sub> = a + 99d
\n\u0c30\u0c46\u0c02\u0c21\u0c35\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f 100\u0c35 \u0c2a\u0c26\u0c02 b100<\/sub> = b + 99d
\n\u0c32\u0c46\u0c15\u0c4d\u0c15\u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02, a100<\/sub> – b100<\/sub> = 100
\n(a + 99d) – (b + 99d) = 100
\na – b = 100 ………….. (1)
\n\u0c07\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41,
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c3f 1000\u0c35 \u0c2a\u0c26\u0c02 a1000<\/sub> = a + 999d
\n\u0c30\u0c46\u0c02\u0c21\u0c35\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f’ 1000\u0c35 \u0c2a\u0c26\u0c02 b10000<\/sub> = b + 999d
\na1000<\/sub> – b1000<\/sub> = (a + 999d) – (b + 999d)
\n= a – b = 100 ((1) \u0c28\u0c41\u0c02\u0c21\u0c3f)
\n\u2234 1000\u0c35 \u0c2a\u0c26\u0c3e\u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c24\u0c47\u0c21\u0c3e 100.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 11.
\n7 \u0c1a\u0c47 \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c2c\u0c21\u0c47 \u0c2e\u0c42\u0c21\u0c02\u0c15\u0c46\u0c32 \u0c38\u0c02\u0c16\u0c4d\u0c2f\u0c32\u0c41 \u0c0e\u0c28\u0c4d\u0c28\u0c3f \u0c15\u0c32\u0c35\u0c41?
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c4d\u0c26\u0c24\u0c3f’:
\n7 \u0c1a\u0c47 \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c2a\u0c2c\u0c21\u0c47 \u0c2e\u0c42\u0c21\u0c02\u0c15\u0c46\u0c32 \u0c38\u0c02\u0c16\u0c4d\u0c2f\u0c32 \u0c1c\u0c3e\u0c2c\u0c3f\u0c24\u0c3e 105, 112, 119, 126, …………., 994 \u0c08 \u0c1c\u0c3e\u0c2c\u0c3f\u0c24\u0c3e \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.
\na = 105; d = a2<\/sub> – a1<\/sub> = 112 – 105 = 7;
\nan<\/sub> = 994
\n\u2234 an<\/sub> = a + (n – 1) d = 994
\n= 105 + (n – 1) 7 = 994
\n105 + 7n – 7 = 994
\n7n + 98 = 994
\n7n = 994 – 98 = 896
\nn = \\(\\frac{896}{7}\\) = 128
\n\u2234 7 \u0c1a\u0c47 \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c2a\u0c2c\u0c21\u0c47 \u0c2e\u0c42\u0c21\u0c02\u0c15\u0c46\u0c32 \u0c38\u0c02\u0c16\u0c4d\u0c2f\u0c32\u0c41 128 \u0c15\u0c32\u0c35\u0c41.<\/p>\n

\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c4d\u0c26\u0c24\u0c3f :
\nd = \\(\\frac{a_{n}-a_{m}}{n-m}\\)
\na1<\/sub> = 105, an<\/sub> = 994, d = 7, m = 1
\n7 = \\(\\frac{994-105}{n-1}=\\frac{896}{n-1}\\)
\nn – 1 = \\(\\frac{889}{7}\\) = 127
\n\u2234 n = 127 +1 = 128.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 12.
\n10 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 250 \u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c17\u0c32 4 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c17\u0c41\u0c23\u0c3f\u0c1c\u0c3e\u0c32 \u0c38\u0c02\u0c16\u0c4d\u0c2f\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n10 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 250 \u0c32 \u0c2e\u0c27\u0c4d\u0c2f \u0c17\u0c32 4 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c17\u0c41\u0c23\u0c3f\u0c1c\u0c3e\u0c32 \u0c1c\u0c3e\u0c2c\u0c3f\u0c24\u0c3e 12, 16, 20, ……… 248.
\n\u0c08 \u0c1c\u0c3e\u0c2c\u0c3f\u0c24\u0c3e A.P \u0c32\u0c4b \u0c15\u0c32\u0c26\u0c41.
\n\u2234 a = 12, d = a2<\/sub> – a1<\/sub> = 16 – 12 = 4,
\nan<\/sub> = 248
\nan<\/sub> = a + (n – 1) d = 248 .
\n= 12 + (n- 1) 4 = 248
\n= 12 + 4n – 4 = 248
\n4n = 248 – 8 = 240
\nn = \\(\\frac{240}{4}\\) = 60
\n\u2234 10 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 250 \u0c32 \u0c2e\u0c27\u0c4d\u0c2f\u0c17\u0c32 4 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c17\u0c41\u0c23\u0c3f\u0c1c\u0c3e\u0c32 \u0c38\u0c02\u0c16\u0c4d\u0c2f = 60.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 13.
\n63, 65, 67, …. \u0c2e\u0c30\u0c3f\u0c2f\u0c41 3, 10, 17, ….. \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32 n\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\u0c2e\u0c4a\u0c26\u0c1f\u0c3f A.P = 63, 65, 67, ……………..
\na = 63, d = a2<\/sub> – a1<\/sub> = 65 – 63 = 2
\n\u2234 n\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = a + (n-1) d
\n= 63 + (n – 1) 2
\n= 63 + 2n – 2
\nn\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = 2n + 61 ………….. (1)
\n\u0c30\u0c46\u0c02\u0c21\u0c35 A.P. = 3, 10, 17, ……………….
\na = 3, d = a2<\/sub> – a1<\/sub> = 10 – 3 = 7
\nn\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = 3 + (n -1 ) 7
\n= 3 + 7n – 7
\nn\u0c35 \u0c2a\u0c26\u0c02 an<\/sub> = 7n – 4 ………… (2)
\n\u0c15\u0c3e\u0c28\u0c3f \u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32 \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c2e\u0c41.
\n\u2234 7n – 4 = 2n + 61
\n7n – 2n = 61 + 4
\n5n = 65
\nn = \\(\\frac{65}{5}\\) = 13
\n\u2234 n = 13.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 14.
\n3\u0c35 \u0c2a\u0c26\u0c2e\u0c41 167; 7\u0c35 \u0c2a\u0c26\u0c2e\u0c41, 5\u0c35 \u0c2a\u0c26\u0c2e\u0c41 \u0c15\u0c02\u0c1f\u0c47 12 \u0c0e\u0c15\u0c4d\u0c15\u0c41\u0c35\u0c17\u0c3e \u0c17\u0c32 \u0c12\u0c15 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c3f\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nA.P \u0c32\u0c4b 3\u0c35 \u0c2a\u0c26\u0c02 a3<\/sub> = a + 2d = 16 ….. (1)
\n5\u0c35 \u0c2a\u0c26\u0c02 a5<\/sub> = a + 4d
\n7\u0c35 \u0c2a\u0c26\u0c02 a7<\/sub> = a + 6d
\n\u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 7\u0c35 \u0c2a\u0c26\u0c2e\u0c41, 5\u0c35 \u0c2a\u0c26\u0c2e\u0c41 \u0c15\u0c02\u0c1f\u0c47 12 \u0c0e\u0c15\u0c4d\u0c15\u0c41\u0c35.
\na + 6d = (a + 4d) + 12
\na + 6d – a – 4d = 12
\n2d = 12 \u21d2 d = \\(\\frac{12}{2}\\) = 6
\nd = 6 \u0c28\u0c41 (1) \u0c32\u0c4b \u0c2a\u0c4d\u0c30\u0c24\u0c3f\u0c15\u0c4d\u0c37\u0c47\u0c2a\u0c3f\u0c02\u0c1a\u0c17\u0c3e,
\na + 2(6) = 16
\na + 12 = 16
\na = 16 – 12 = 4
\na = 4 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 d = 6
\n\u2234 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f 4, 10, 16, 22, …………….<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 15.
\n3, 8, 13, ….., 253 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c1a\u0c3f\u0c35\u0c30\u0c3f \u0c28\u0c41\u0c02\u0c1a\u0c3f 20\u0c35 \u0c2a\u0c26\u0c2e\u0c41\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c4d\u0c26\u0c24\u0c3f :
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P = 3, 8, 13, ………., 253
\n\u0c07\u0c15\u0c4d\u0c15\u0c21 a = 3, d = a2<\/sub> – a1<\/sub> = 8 – 3 = 5,
\n\u21d2an<\/sub> = 253
\n\u21d2 an<\/sub> = a + (n – 1) 4 = 253.
\n\u21d2 3 + (n- 1) 5 = 253
\n\u21d2 3 + 5n – 5 = 253
\n\u21d2 5n = 253 + 2 = 255
\n\u21d2 n = \\(\\frac{255}{5}\\) = 51
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P \u0c32\u0c4b 51 \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c15\u0c32\u0c35\u0c41.
\n\u2234 \u0c1a\u0c3f\u0c35\u0c30\u0c3f \u0c28\u0c41\u0c02\u0c21\u0c3f 20\u0c35 \u0c2a\u0c26\u0c02, \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c28\u0c41\u0c02\u0c21\u0c3f (51 – 20) + 1 = 32\u0c35 \u0c2a\u0c26\u0c02 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.
\n\u2234 32\u0c35 \u0c2a\u0c26\u0c02 a32<\/sub> = 3 + (32 – 1) (5)
\n= 3 + 31 (5)
\na32<\/sub> = 3 + 155 = 158
\n\u0c1a\u0c3f\u0c35\u0c30\u0c3f \u0c28\u0c41\u0c02\u0c21\u0c3f 20\u0c35 \u0c2a\u0c26\u0c02 = 158<\/p>\n

2\u0c35 \u0c2a\u0c26\u0c4d\u0c26\u0c24\u0c3f :
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 A.P = 3, 8, 13, ….., 253
\n\u0c07\u0c15\u0c4d\u0c15\u0c21 d = a2<\/sub> – a1<\/sub> = 8 – 3 = 5
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c28\u0c3f \u0c24\u0c4d\u0c30\u0c3f\u0c2a\u0c4d\u0c2a\u0c3f \u0c30\u0c3e\u0c2f\u0c17\u0c3e \u0c35\u0c1a\u0c4d\u0c1a\u0c47 20\u0c35 \u0c2a\u0c26\u0c2e\u0c47 \u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c3f \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c1a\u0c3f\u0c35\u0c30\u0c3f \u0c28\u0c41\u0c02\u0c21\u0c3f 20\u0c35 \u0c2a\u0c26\u0c02 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f. 253, 248, 243, ………., 13, 8, 3
\n\u0c08 \u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b a = 253, d = a2<\/sub> – a1<\/sub>
\n= 248 – 253 = – 5
\nan<\/sub> = 3
\nan<\/sub> = a + (n – 1) d = 3
\n253 + (n – 1) (- 5) = 3
\n253 – 5n + 5 = 3
\n258 – 5n = 3
\n– 5n = 3 – 258 = – 255
\n5n = 255
\n\u21d2 n = \\(\\frac{255}{5}\\) = 51
\n\u2234 20\u0c35 \u0c2a\u0c26\u0c02 a20<\/sub> = 253 + (20 – 1) (- 5)
\n= 253 – 95
\nan<\/sub> = 158
\n\u2234 3, 8, 13, …………. 253 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c22\u0c3f \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c1a\u0c3f\u0c35\u0c30\u0c3f \u0c28\u0c41\u0c02\u0c21\u0c3f 20\u0c35 \u0c2a\u0c26\u0c2e\u0c41 = 158.<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 16.
\n\u0c12\u0c15 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b 4\u0c35 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 8\u0c35 \u0c2a\u0c26\u0c3e\u0c32 \u0c2e\u0c4a\u0c24\u0c4d\u0c24\u0c2e\u0c41 24 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 6\u0c35, 10\u0c35 \u0c2a\u0c26\u0c3e\u0c32 \u0c2e\u0c4a\u0c24\u0c4d\u0c24\u0c2e\u0c41 44 \u0c05\u0c2f\u0c3f\u0c28 \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2e\u0c42\u0c21\u0c41 \u0c2a\u0c26\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\n\u0c38\u0c3e\u0c27\u0c28.
\nA.P \u0c32\u0c4b 4\u0c35 \u0c2a\u0c26\u0c02 = a + 3d
\n8\u0c35 \u0c2a\u0c26\u0c02 = a + 7d
\n\u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 4\u0c35, 8\u0c35 \u0c2a\u0c26\u0c3e\u0c32 ,\u0c2e\u0c4a\u0c24\u0c4d\u0c24\u0c02 = 24
\n(a + 3d) + (a + 7d) = 24
\n= 2a + 10d = 24
\n2 (a + 5d) = 24
\na + 5d = \\(\\frac{24}{2}\\) = 12
\n\u2234 a + 5d = 12 ………….. (1)
\n\u0c07\u0c32\u0c3e\u0c17\u0c47, 6\u0c35 \u0c2a\u0c26\u0c02 = a + 5d
\n10\u0c35 \u0c2a\u0c26\u0c02 = a + 9d
\n6\u0c35 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 10\u0c35 \u0c2a\u0c26\u0c3e\u0c32 \u0c2e\u0c4a\u0c24\u0c4d\u0c24\u0c02 44
\n(a + 5d) + (a + 9d) = 44
\n2a + 140 = 44
\n2 (a + 7d) = 44
\na + 7d = \\(\\frac{44}{2}\\) = 22
\na + 7d = 22 ………… (2)
\n(2) – (1)<\/p>\n

\"AP<\/p>\n

d = \\(\\frac{10}{2}\\) = 5
\n\u2234 d = 2
\nd = 5\u0c28\u0c41 (1) \u0c32\u0c4b \u0c2a\u0c4d\u0c30\u0c24\u0c3f\u0c15\u0c4d\u0c37\u0c47\u0c2a\u0c3f\u0c02\u0c1a\u0c17\u0c3e,
\na + 5(5) = 12 \u21d2 a = 12 – 25 = – 13
\n\u2234 \u0c15\u0c3e\u0c35\u0c32\u0c38\u0c3f\u0c28 \u0c05\u0c02\u0c15\u0c36\u0c4d\u0c30\u0c47\u0c23\u0c3f\u0c32\u0c4b\u0c28\u0c3f \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2e\u0c42\u0c21\u0c41 \u0c2a\u0c26\u0c3e\u0c32\u0c41
\n\u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 a1<\/sub> = a = – 13
\n\u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 a2<\/sub> = – 13 + 5 = – 8
\n\u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c02 a3<\/sub> = 3 – 8 + 5 = – 3<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 17.
\n\u0c38\u0c41\u0c2c\u0c4d\u0c2c\u0c3e\u0c30\u0c3e\u0c35\u0c41 1995\u0c35 \u0c38\u0c02||\u0c32\u0c4b \u0c28\u0c46\u0c32\u0c15\u0c41 \u20b9 5000 \u0c1c\u0c40\u0c24\u0c02\u0c24\u0c4b \u0c09\u0c26\u0c4d\u0c2f\u0c4b\u0c17\u0c02\u0c32\u0c4b \u0c1a\u0c47\u0c30\u0c3e\u0c21\u0c41. \u0c05\u0c24\u0c28\u0c3f \u0c1c\u0c40\u0c24\u0c2e\u0c41 \u0c38\u0c02||\u0c2e\u0c41\u0c28\u0c15\u0c41 \u20b9 200 \u0c2a\u0c46\u0c30\u0c3f\u0c17\u0c3f\u0c28 \u0c05\u0c24\u0c28\u0c3f \u0c1c\u0c40\u0c24\u0c2e\u0c41 \u0c0f \u0c38\u0c02||\u0c2e\u0c41\u0c32\u0c4b \u20b9 7000 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f ?
\n\u0c38\u0c3e\u0c27\u0c28.<\/p>\n

\"AP<\/p>\n

\u0c1c\u0c40\u0c24\u0c02 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c1c\u0c3e\u0c2c\u0c3f\u0c24\u0c3e
\n5000, 5200, 5400, 5600, ………….
\n\u0c08 \u0c1c\u0c3e\u0c2c\u0c3f\u0c24\u0c3e A.P \u0c32\u0c4b \u0c15\u0c32\u0c26\u0c41.
\n\u2234 a = 5000, d = a2<\/sub> – a1<\/sub> = 5200 – 5000 = 200
\nan<\/sub> = 7000
\nan<\/sub> = a + (n – 1) 4 = 7000
\n= 5000 + (n – 1) 200 = 7000
\n= 5000 + 200 n – 200 = 7000
\n200 n = 7000 – 4800 = 2200
\n\u2234 n = \\(\\frac{2200}{200}\\) = 11.
\n\u0c1c\u0c3e\u0c2c\u0c3f\u0c24\u0c3e\u0c32\u0c4b 7000 11\u0c35 \u0c2a\u0c26\u0c02 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.
\n\u0c05\u0c28\u0c17\u0c3e ,\u0c38\u0c41\u0c2c\u0c4d\u0c2c\u0c3e\u0c30\u0c3e\u0c35\u0c41 \u0c09\u0c26\u0c4d\u0c2f\u0c4b\u0c17\u0c02\u0c32\u0c4b \u0c1a\u0c47\u0c30\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c1f\u0c3f \u0c28\u0c41\u0c02\u0c21\u0c3f 11\u0c35 \u0c38\u0c02||\u0c32\u0c4b \u0c05\u0c24\u0c28\u0c3f \u0c1c\u0c40\u0c24\u0c02 \u20b9 7000 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f. (1995\u0c28\u0c41 \u0c15\u0c42\u0c21\u0c3e \u0c15\u0c32\u0c41\u0c2a\u0c41\u0c15\u0c4b\u0c35\u0c3e\u0c32\u0c3f)
\n\u2234 2005 \u0c35 \u0c38\u0c02||\u0c32\u0c4b \u0c38\u0c41\u0c2c\u0c4d\u0c2c\u0c3e\u0c30\u0c3e\u0c35\u0c41 \u0c1c\u0c40\u0c24\u0c02 \u20b9 7000 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions Chapter 6 \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.2 Textbook Exercise Questions and Answers. AP State Syllabus 10th Class Maths Solutions 6th Lesson \u0c36\u0c4d\u0c30\u0c47\u0c22\u0c41\u0c32\u0c41 Exercise 6.2 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c2e\u0c41 a, \u0c38\u0c3e\u0c2e\u0c3e\u0c28\u0c4d\u0c2f \u0c2d\u0c47\u0c26\u0c2e\u0c41 d, n\u0c35 \u0c2a\u0c26\u0c2e\u0c41 a, \u0c05\u0c2f\u0c3f\u0c28 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2a\u0c1f\u0c4d\u0c1f\u0c3f\u0c15\u0c28\u0c41 \u0c2a\u0c42\u0c30\u0c3f\u0c02\u0c2a\u0c41\u0c2e\u0c41. – AS,, AS,, \u0c38\u0c3e\u0c27\u0c28. \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2. \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41. (i) 10, …<\/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\/27354"}],"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=27354"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/27354\/revisions"}],"predecessor-version":[{"id":27378,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/27354\/revisions\/27378"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=27354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=27354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=27354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}