how many terms of the ap 3,7,11,15,.... will make the sum 406?
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Given,
AP:- 3, 7, 11, 15 ......
Sn=406
To find,
Value of n
ATQ,
Here,
T1=3, T2=7, d=T2-T1=7-3=4
Now,
Sn=n/2[2a+(n-1)d]
406=n/2[2*3+(n-1)4]
406=n/2[6+4n-4]
406=n/2[4n+2]
406=(4n^2+2n)/2
0=2n^2+n-406
2(n^2+n/2-203)=0
(n)^2+2*n*1/4+(1/4)^2-(1/4)^2-203=0
(n+1/4)^2-1/16-203=0
(n+1/4)^2=(1+3248)/16
(n+1/4)=√(3249/16)
n+1/4=57/4
n=(57-1)/4
n=56/4
n=14
Hope it helps you!
AP:- 3, 7, 11, 15 ......
Sn=406
To find,
Value of n
ATQ,
Here,
T1=3, T2=7, d=T2-T1=7-3=4
Now,
Sn=n/2[2a+(n-1)d]
406=n/2[2*3+(n-1)4]
406=n/2[6+4n-4]
406=n/2[4n+2]
406=(4n^2+2n)/2
0=2n^2+n-406
2(n^2+n/2-203)=0
(n)^2+2*n*1/4+(1/4)^2-(1/4)^2-203=0
(n+1/4)^2-1/16-203=0
(n+1/4)^2=(1+3248)/16
(n+1/4)=√(3249/16)
n+1/4=57/4
n=(57-1)/4
n=56/4
n=14
Hope it helps you!
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