Posted by akash on .
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.

arithmetic 
Steve,
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 
arithmetic 
Sebin PS,
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 
Trigonometric Function 
7.1,
2ab/a² b²