arithmetic
posted by akash .
In an AP the first term is 2 and the sum of the first five terms is one fourth of the sum of the next five terms. Show that the 20th term is 112.

nth term is a+(n1)d
sum of 1st n terms = n/2(T1+Tn)
a = 2
sum of 1st 5 terms = 5/2 (2 + 2+4d) = 10+10d
6th term = 2+5d
sum of terms 610 = 5/2 (2+5d + 2+9d) = 10+35d
10+10d = 1/4 (10+35d)
40+40d = 10 + 35d
5d = 30
d = 6
sequence is 2 4 10 16 22 28 34 40 46 52
sum of T1..T5 = 50
sum of T6..T10 = 200 
nth term is a+(n1)d
sum of 1st n terms = n/2(T1+Tn)
a = 2
sum of 1st 5 terms = 5/2 (2 + 2+4d) = 10+10d
6th term = 2+5d
sum of terms 610 = 5/2 (2+5d + 2+9d) = 10+35d
10+10d = 1/4 (10+35d)
40+40d = 10 + 35d
5d = 30
d = 6
sequence is 2 4 10 16 22 28 34 40 46 52
sum of T1..T5 = 50
sum of T6..T10 = 200
(copied From above)
20th Term = a +(n1)d
= 2+19 X 6
= 2  114
= 112 
2ab/a² b²