অনুশীলনী – 5.2
1. দিয়া আছে যে সমান্তৰ প্ৰগতিৰ প্ৰথম পদ \(a_1\) সাধাৰণ অন্তৰ d আৰু n তম পদ \(a_n\) । তলৰ তালিকাখনৰ খালী ঠাইসমূহ পূৰণ কৰা 一
| \(a\) | d | n | \(a_n\) |
| 7 | 3 | 8 | |
| -18 | 10 | 0 | |
| -3 | 18 | -5 | |
| -18.9 | 2.5 | 3.6 | |
| 3.5 | 0 | 105 |
(i) দিয়া আছে,
\(a\) = 7
d = 3
n = 8
আমি জানো যে,
⇒ \(a_n = a + (n-1)d\)
⇒ \(a_n = 7 + (8-1)3\) = 7 + 21 = 28
(ii) দিয়া আছে,
\(a\) = -18
\(a_n\) = 0
n = 10
আমি জানো যে,
⇒ \(a_n = a + (n-1)d\)
⇒ \( 0 = -18 + (10-1)d\)
⇒ 18 = 9d
⇒ d = 2
(iii) দিয়া আছে,
d = -3
\(a_n\) = -5
n = 18
আমি জানো যে,
⇒ \(a_n = a + (n-1)d\)
⇒ \( -5 = a + (18-1)(-3)\)
⇒ \(-5 = a – 51\)
⇒ a = -5 +51 = 46
(iv) দিয়া আছে,
\(a\) = -18.9
\(d\) = 2.5
\(a_n\) = 3.6
আমি জানো যে,
⇒ \(a_n = a + (n-1)d\)
⇒ \( 3.6 = -18.9 + (n-1)2.5\)
⇒ 3.6 + 18.9 = 2.5(n-1)
⇒ n-1 = \(\frac{22.5}{2.5}\) = 9
⇒ n = 9 + 1 =10
(v) দিয়া আছে,
\(a\) = 3.5
d = 0
n = 105
আমি জানো যে,
⇒ \(a_n = a + (n-1)d\)
⇒ \(a_n = 3.5 + (105-1)0\) = 3.5
2. তলৰ প্ৰতিটোৰে শুদ্ধ উত্তৰটো বাছি উলিওৱা আৰু কাৰণ দৰ্শোৱা 一
(i) 10, 7, 4, ……., এই সমান্তৰ প্ৰগতিটোৰ 30 তম পদটো
(a) 97 (B) 77 (C) -77 (D) -87
সমাধানঃ- (i) দিয়া আছে,
\(a\) = 10
d = 7 – 10 = -3
n = 30
আমি জানো যে,
⇒ \(a_{30} = 10 + (30-1)(-3)\)
⇒ \(a_{30} = 10 + (29)(-3)\) = 10 – 87 = -77
উত্তৰঃ (C) -77
(ii) -3, -1/2, 2, ……., এই সমান্তৰ প্ৰগতিটোৰ 11তম পদটো
(A) 28 (B) 22 (C) -38 (D) -48 1/2
সমাধানঃ- (ii) দিয়া আছে,
\(a\) = -3
⇒ d = \(\frac{-1}{2} – (-3)\) = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\)
⇒ n = 30
আমি জানো যে,
⇒ \(a_{11} = -3 + (11-1)\) \(\frac{5}{2}\)
⇒ \(a_{11} = -3 + 10\)×\(\frac{5}{2}\) = -3 + 25 = 22
উত্তৰঃ (B) 22
3. তলৰ সমান্তৰ প্ৰগতিসমূহৰ খালীঘৰ কেইটাৰ লুপ্ত পদসমূহ নিৰ্ণয় কৰা —-
(i) 2, \(\square\) , 26
Solution:-
দিয়া আছে,
\(a_1 = 2\)
এতেকে, \(a = 2\)
⇒ \(a_3 = 26\)
⇒ \(a + 2d = 26\)
⇒ \(2+ 2d = 26\)
⇒ \(2d = 26 – 2\)
⇒ \(2d = 24\)
⇒ \(d = 12\)
(ii) \(\square\) , 13 , \(\square\) , 3
Solution:-
Given, \(a_2 = 13\)
\(a_4 = 3\)
এতেকে,
⇒ \(a_2 = 13\)
⇒ \(a + d = 13\) ______________________ (i)
আৰু,
⇒ \(a_4 = 3\)
⇒ \(a + 3d = 3\)
⇒ \(a + d + 2d = 3\)
(i) ৰ মান বহুৱাই
⇒ 13 + 2d =3
⇒ 2d = 3 -13
⇒ 2d = -10
⇒ d = -5
d ৰ মান (i) ত বহুৱাই
⇒ \(a + (-5) = 13\)
⇒ \(a = 18\)
⇒ \(a_3 = a + 2d = 18 + 2(-5) \) =8
(iii) 5 , \(\square\) , \(\square\) , \(9 \frac{1}{2}\)
Solution:-
Given,
\(a = 5\)
\(a_4 = 9 \frac{1}{2}\)
এতেকে,
⇒ \(a + (4-1)d = 9 \frac{1}{2}\)
⇒ \(5 + 3d = \frac{19}{2}\)
⇒ \(3d = \frac{19}{2} – 5\)
⇒ \(3d = \frac{19-10}{2}\)
⇒ \(d = \frac{9}{6}\)
⇒ \(a_2 = a + d\) = \(5 + \frac{9}{6}\) = \(5 + \frac{3}{2}\) = \(\frac{10+3}{2}\) = \(\frac{13}{2}\)
⇒ \(a_3 = a + 2d\) = \(5 + 2\frac{9}{6}\) =\(5 + \frac{9}{3}\) = \(\frac{15+9}{3}\) = \(\frac{24}{3}\) = 8
(iv) -4 , \(\square\) , \(\square\) , \(\square\) , \(\square\) ,6
Solution:-
Given,
\(a = -4\)
\(a_6 = 6\)
এতেকে,
⇒ \(a _6 = 6\)
⇒ \(-4 + 5d = 6\)
⇒ \(5d = 6 + 4\)
⇒ \(5d = 10 \)
⇒ \(d = 2\)
⇒ \(a_2 = a + d\) = \(-4 + 2\) = -2
⇒ \(a_3 = a + 2d\) = \(-4 + 2(2)\) =\(-4 + 4\) = 0
⇒ \(a_4 = a + 3d\) = \(-4 + 3(2)\) = -4 + 6 = 2
⇒ \(a_5 = a + 4d\) = \(-4 + 4(2)\) =\(-4 + 8\) = 4
(v) \(\square\) , 38 ,\(\square\) , \(\square\) , \(\square\) , -22
Solution:-
Given,
\(a_2 = 38\)
\(a_6 = -22\)
এতেকে,
⇒ \(a _2 = 38\)
⇒ \(a + d = 38\) ——————— (i)
and ,
⇒ \(a_6 = -22\)
⇒ \(a + 5d = -22\)
⇒ \(a + d + 4d= -22\)
⇒ \(38 + 4d = -22\)
⇒ \(4d = -22 – 38\)
⇒ \(4d = -60\)
⇒ \(d = – 15\)
⇒ \(a + d = 38\)
⇒ \(a + (-15) = 38\)
⇒ \(a = 38 + 15 = 53\)
⇒ \(a_3 = a + 2d = 53 +2(-15) = 53 – 30 = 23\)
⇒ \(a_4 = a + 3d = 53 + 3(-15) = 53 – 45 = 8\)
⇒ \(a_5 = a + 4d = 53 +4(-15) = 53 – 60 = -7\)