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.