Question

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

Answer 1

(i) Given, \(a = 2\) and \(a_3 = 26\)
We know that, \(a_n = a + (n − 1) d\)
\(a_3 = 2 + (3 − 1) d\)
\(26 = 2 + 2d\)
\(24 = 2d\)
\(d = 12\)
\(a_2 = 2 + (2 − 1) 12\)
\(a_2= 14\)

Therefore, 14 is the missing term.

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(ii) Given, \(a_2 = 13\) and \(a_4 = 3\)
We know that,
\(a_n = a + (n − 1) d\)
\(a_2 = a + (2 − 1) d\)
\(13 = a + d\)      ……(I)
\(a_4 = a + (4 − 1) d\)
\(3= a + 3d\)       ……..(II)
On subtracting (I) from (II), we obtain
\(−10 = 2d\)
\(d = −5\)
From equation (I), we obtain
\(13 = a + (−5)\)
\(a = 18\)
\(a_3 = 18 + (3 − 1) (−5)\)
\(a_3 = 18 + 2 (−5)\)
\(a_3 = 18 − 10\)
\(a_3= 8\)

Therefore, the missing terms are 18 and 8 respectively.

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(iii) Given, \(a = 5\) and \(a_4 = 9\frac {1}{2} =\frac {19}{2}\)
We know that,
\(a_n = a + (n-1)d\)
\(a_4 = a + (4-1)d\)
\(\frac {19}{2} = 5 + 3d\)

\(\frac {19}{2} - 5 = 3d\)

\(\frac {9}{2} = 3d\)

\(d = \frac 32\)
\(a_2 = a+d = 5 + \frac 32 = \frac {13}{2}\)
\(a_3 = a+2d = 5 + 2(\frac 32) = 8\)

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

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(iv) Given, \(a = −4\) and \(a_6 = 6\)
We know that,
\(a_n = a + (n − 1) d\)
\(a_6 = a + (6 − 1) d\)
\(6 = − 4 + 5d\)
\(10 = 5d\)
\(d = 2\)
\(a_2 = a + d = − 4 + 2 = −2\)
\(a_3 = a + 2d = − 4 + 2 \times 2 = 0\)
\(a_4 = a + 3d = − 4 + 3 \times 2 = 2\)
\(a_5 = a + 4d = − 4 + 4 \times 2 = 4\)

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

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(v) Given, \(a_2 = 38\) and \(a_6 = −22\)
We know that;
\(a_n = a + (n − 1) d\)
\(a_2 = a + (2 − 1) d\)
\(38 = a + d\)      ………(1)
\(a_6 = a + (6 − 1) d\)
\(−22 = a + 5d\)     ……….(2)
On subtracting equation (1) from (2), we obtain
\(− 22 − 38 = 4d\)
\(−60 = 4d\)
\(d = −15\)
\(a = a_2 − d = 38 − (−15) = 38 +15= 53\)
\(a_3 = a + 2d = 53 + 2 (−15) = 53-30= 23\)
\(a_4 = a + 3d = 53 + 3 (−15) = 53-45 = 8\)
\(a_5 = a + 4d = 53 + 4 (−15) = 53-60= −7\)

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