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

(i) For this A.P.,a=2a3=26We know that,an=a+(n−1;)da3=2+(3−1;)d26=2+2d24=2dd=12a2=2+(2−1;)12=14Therefore,14is the missing term (ii) For this A.P.,a2=13anda4=3We know that,an=a+(n−1;)da2=a+(2−1;)d13=a+d... (i)a4=a+(4−1;)d3=a+3d... (ii)On subtracting (i) from (ii), we get,−1;0=2dd=−5From equation (i), we get,13=a+(−5)a=18a3=18+(3−1;)(−5)=18+2(−5)=18−1;0=8Therefore, the missing terms are18and8respectively (iii) For this A.P.,a1=5anda4=912We know that,an=a+(n−1;)da4=5+(4−1;)d912=5+3dd=32a2=a+da2=5+32a2=132a3=a2+32a3=8Therefore, the missing terms are612and8respectively (iv) For this A.P.,a=−4anda6=6We know that,an=a+(n−1;)da6=a+(6−1;)d6=−4+5d10=5dd=2a2=a+d=−4+2=−2;a3=a+2d=−4+2(2)=0a4=a+3d=−4+3(2)=2a5=a+4d=−4+4(2)=4Therefore, the missing terms are−2;,0,2,and4respectively (v) For this A.P.,a2=38a6=−2;2We know thatan=a+(n−1;)da2=a+(2−1;)d38=a+d... (i)a6=a+(6−1;)d−2;2=a+5d... (ii)On subtracting equation (i) from (ii), we get−2;2−38=4d−6;0=4dd=−1;5a=a2−a=38−(−1;5)=53a3=a+2d=53+2(−1;5)=23a4=a+3d=53+3(−1;5)=8a5=a+4d=53+4(−1;5)=−7Therefore, the missing terms are53,23,8and−7respectively.