Question

find the sum of first 20 terms of an AP whose nth term is 4n -1

Answer 1

given
an=4n-1
a1=4*1-1=4-1=3
a2=4*2-1=8-1=7
a3=4*3-1=12-1=11
therefore
first term =a1=a=3
common difference=d=a2-a1=7-3=4
sum of n terms =sn
sn=n/2[2a+(n-1)d]
=n/2[2*3+(n-1)4]
=n/2[6+4n-4]
=n[3+2n-2]
=n[2n+1]----(1)
if n=20
s20=20[2*20+1]
=20*41
=820

Answer 2

Tn = 4n-1
-----------------
If n = 1 ;

T1 = 4(1)-1 = 3
a = 3
----------------
If n = 2 ;

T2 = 4(2)-1 = 7
a+d = 7
---------------

3 = 7-d
d = 4
=============
Now, in a A.P
a = 3
d = 4
n = 20
S20 = ?

Sn = (n/2)[2a+(n-1)d]

S20 = (20/2) [(2×3)+(20-1)4]
= 10 ×(6+76)
= 10×82
= 820
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