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Question

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

(i) For this A.P.,
a=2
a3=26
We know that, an=a+(n1)d
a3=2+(31)d
26=2+2d
24=2d
d=12
a2=2+(21)12
=14
Therefore, 14 is the missing term.


(ii) For this A.P.,
a2=13 and
a4=3
We know that, an=a+(n1)d
a2=a+(21)d
13=a+d ... (i)
a4=a+(41)d
3=a+3d... (ii)
On subtracting (i) from (ii), we get,
10=2d
d=5
From equation (i), we get,
13=a+(5)
a=18
a3=18+(31)(5)
=18+2(5)=1810=8
Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,
a1=5 and
a4=912
We know that, an=a+(n1)d
a4=5+(41)d
912=5+3d
d=32

a2=a+d
a2=5+32
a2=132
a3=a2+32
a3=8
Therefore, the missing terms are 612 and 8 respectively.

(iv) For this A.P.,
a=4 and
a6=6
We know that,
an=a+(n1)d
a6=a+(61)d
6=4+5d
10=5d
d=2
a2=a+d=4+2=2
a3=a+2d=4+2(2)=0
a4=a+3d=4+3(2)=2
a5=a+4d=4+4(2)=4
Therefore, the missing terms are 2,0,2, and 4 respectively.

(v) For this A.P.,
a2=38
a6=22
We know that
an=a+(n1)d
a2=a+(21)d
38=a+d... (i)
a6=a+(61)d
22=a+5d ... (ii)
On subtracting equation (i) from (ii), we get
2238=4d
60=4d
d=15
a=a2a=38(15)=53
a3=a+2d=53+2(15)=23
a4=a+3d=53+3(15)=8
a5=a+4d=53+4(15)=7
Therefore, the missing terms are 53,23,8 and 7 respectively.

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