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Question

How many terms of the AP 3,7,11,15,... willl make the sum 406 ?
(a) 10 (b) 12 (c) 14 (d) 20

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Solution

AP:3,7,11,15,..a=3d=73=4Sn=406We know that,Sn=n2[2a+(n1)d]n2[2(3)+(n1)4]=406n2[6+4n4]=406n2[4n+2]=406n[2n+1]=4062n2+n406=02n228n+29n406=02n(n14)+29(n14)=0(2n+29)(n14)=0n=292 or 14

n cannot be negative.

So n = 14

Hence, 14 terms will make the sum 406.


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