if the nth term of a series is (3+n)/4 then the sum of 105 terms of the series is
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first put n=1. secondly put n=2
(3+n)/4. (3+2)/4
(3+1)/4. 5/4
4/4.
1
a=1.
d=T2-T1
d= 5/4 -1
d=1/4
number of term= 105
![sn = n \div 2(2a + (n - 1)d) \\ sn = 105 \div 2(2 \times + 1(105 - 1)1 \div 4) \\ sn = n \div 2(2a + (n - 1)d) \\ sn = 105 \div 2(2 \times + 1(105 - 1)1 \div 4) \\](https://tex.z-dn.net/?f=sn+%3D+n+%5Cdiv+2%282a+%2B+%28n+-+1%29d%29+%5C%5C+sn+%3D+105+%5Cdiv+2%282+%5Ctimes++%2B+1%28105+-+1%291+%5Cdiv+4%29+%5C%5C)
Sn=105/2(2+104*1/4)
Sn=105 /2(2+26)
Sn=105/2*28
Sn=105*14
Sn=1470
(3+n)/4. (3+2)/4
(3+1)/4. 5/4
4/4.
1
a=1.
d=T2-T1
d= 5/4 -1
d=1/4
number of term= 105
Sn=105/2(2+104*1/4)
Sn=105 /2(2+26)
Sn=105/2*28
Sn=105*14
Sn=1470
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