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.

Answer 1

nth term is a+(n-1)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 6-10 = 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

Answer 2

nth term is a+(n-1)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 6-10 = 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 +(n-1)d

= 2+19 X -6
= 2 - 114
= -112

Answer 3

2ab/a² b²