Question

In the following APs, find the missing terms in the boxes :

Answer 1

(i) Given: \(a = 2\) and \(a_3 = 26\). Using the formula \(a_n = a + (n − 1) d\), we get \(a_3 = 2 + (3 − 1) d\), which simplifies to \(26 = 2 + 2d\). Solving for \(d\), we find \(2d = 24\), so \(d = 12\). The second term is \(a_2 = 2 + (2 − 1) \times 12 = 2 + 12 = 14\).

Therefore, 14 is the missing term.

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(ii) Given: \(a_2 = 13\) and \(a_4 = 3\). Using the formula \(a_n = a + (n − 1) d\), we have two equations: \(13 = a + d\) (Equation I) and \(3 = a + 3d\) (Equation II). Subtracting Equation I from Equation II gives \(-10 = 2d\), so \(d = -5\). Substituting \(d = -5\) into Equation I, we get \(13 = a + (-5)\), so \(a = 18\). The third term is \(a_3 = 18 + (3 − 1) \times (-5) = 18 + 2 \times (-5) = 18 - 10 = 8\).

Therefore, the missing terms are 18 and 8 respectively.

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(iii) Given: \(a = 5\) and \(a_4 = 9\frac{1}{2} = \frac{19}{2}\). Using the formula \(a_n = a + (n-1)d\), we have \(a_4 = a + (4-1)d\). Substituting the given values, we get \(\frac{19}{2} = 5 + 3d\). Rearranging, we have \(3d = \frac{19}{2} - 5 = \frac{9}{2}\), so \(d = \frac{3}{2}\). The second term is \(a_2 = a + d = 5 + \frac{3}{2} = \frac{13}{2}\). The third term is \(a_3 = a + 2d = 5 + 2(\frac{3}{2}) = 5 + 3 = 8\).

Therefore, the missing terms are \(\frac{13}{2}\) and 8 respectively.

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(iv) Given: \(a = -4\) and \(a_6 = 6\). Using the formula \(a_n = a + (n - 1) d\), we have \(a_6 = a + (6 - 1) d\). Substituting the given values, we get \(6 = -4 + 5d\). Solving for \(d\), we find \(5d = 10\), so \(d = 2\). The missing terms are: \(a_2 = a + d = -4 + 2 = -2\), \(a_3 = a + 2d = -4 + 2 \times 2 = 0\), \(a_4 = a + 3d = -4 + 3 \times 2 = 2\), and \(a_5 = a + 4d = -4 + 4 \times 2 = 4\).

Therefore, the missing terms are -2, 0, 2, and 4 respectively.

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(v) Given: \(a_2 = 38\) and \(a_6 = -22\). Using the formula \(a_n = a + (n - 1) d\), we have two equations: \(38 = a + d\) (Equation 1) and \(-22 = a + 5d\) (Equation 2). Subtracting Equation 1 from Equation 2 gives \(-22 - 38 = 4d\), which simplifies to \(-60 = 4d\), so \(d = -15\). To find \(a\), we use \(a = a_2 - d = 38 - (-15) = 38 + 15 = 53\). The missing terms are: \(a_3 = a + 2d = 53 + 2(-15) = 53 - 30 = 23\), \(a_4 = a + 3d = 53 + 3(-15) = 53 - 45 = 8\), and \(a_5 = a + 4d = 53 + 4(-15) = 53 - 60 = -7\).

Therefore, the missing terms are 53, 23, 8, and -7 respectively.