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In the following APs, find the missing terms in the boxes:
(i) $2, \square, 26$
(ii) $\square, 13, \square, 3$
(iii) $5, \square, \square, 9\frac{1}{2}$
(iv) $-4, \square, \square, \square, \square, 6$
(v) $\square, 38, \square, \square, \square, -22$
To do:
We have to find the missing term in the box in each case.
Solution:
(i) Given AP is $2, \square, 26$
$a=2, a_{3}=26$
We know that,
$a_{n}=a+(n-1) d$
$26=a+2 d$
$26 =2+2 d$
$2 d=26-2$
$2d=24$
$d=\frac{24}{2}$
$d=12$
$a_{2}=a+(2-1) d$
$=a+d$
$=2+12$
$=14$
The missing term in the box is 14.
(ii) Given AP is $\square, 13, \square, 3$
$a=13$
We know that,
$a_{2}=a+(2-1) d$
$a+d=13$......(i)
$a_{4}=3$
$a_{4}=a+(4-1) d$
$a+3 d=3$.........(ii)
Subtracting (i) from (ii), we get,
$a+3d-a-d=3-13$
$2d=-10$
$d=\frac{-10}{2}$
$d=-5$
This implies,
$a+d =13$
$a+(-5)=13$
$a=13+5$
$a=18$
$a_{3}=a+2 d$
$=18+2(-5)$
$=18-10$
$=8$
(iii) Given AP is $5, \square, \square, 9\frac{1}{2}$
$a=5$
$a_4=9\frac{1}{2}$
We know that,
$a_{4}=a+(4-1) d$
$5+3 d=\frac{19}{2}$
$3d=\frac{19}{2}-5$
$3d=\frac{19-10}{2}$
$3d=\frac{9}{2}$
$d=\frac{9}{2(3)}$
$d=\frac{3}{2}$
This implies,
$a_{2}=a+(2-1) d$
$=a+d$
$=5+\frac{3}{2}$
$=\frac{13}{2}$
$a_{3}=a_{2}+d$
$=\frac{13}{2}+\frac{3}{2}$
$=\frac{16}{2}$
$=8$
(iv) Given AP is $-4, \square, \square, \square, \square, 6$
$a=-4, a_{6}=6$
We know that,
$a_{n}=a+(n-1) d$
$a_{2}=a+d$
$=-4+2$
$=-2 $
$a_{3}=a_{2}+d$
$=-2+2$
$=0$
$a_{4}=a_{3}+d$
$=0+2$
$=2$
$a_{5}=a_{4}+d$
$=2+2$
$=4$
(v) Given AP is $\square, 38, \square, \square, \square, -22$
$a_{2}=38, a_{6}=-22$
We know that,
$a_{n}=a+(n-1) d$
$a_{2}=a+d$
$a+d=38$......(i)
$a_{6}=a+5 d$
$=-22$......(ii)
Subtracting (i) from (ii), we get,
$a+5 d-a-d=-22-38$
$4d=-60$
$d=\frac{-60}{4}$
$d=-15$
This implies,
$a+d=38$
$a=38-(-15)$
$a=38+15$
$a=53$
$a_3=a_2+d$
$=38+(-15)$
$=38-15$
$=23$
$a_4=a_3+d$
$=23+(-15)$
$=23-15$
$=8$
$a_5=a_4+d$
$=8+(-15)$
$=8-15$
$=-7$