Question

In an A.P,the first term is 2 and the sum of the first five terms is one-fourth of the next five terms .Show that 20th term is -112.

Answer 1

given; a = 2
A.T.Q
[a1 + a2 + a3 + a4 + a5] = 1/4 [ a6 + a7 + a8 + a9 + a10]
{(a) + (a + d) + (a + 2d) + (a + 3d) + (a + 4d)} = 1/4 {(a+5d) + (a+6d) + (a+7d) + (a+8d) + (a+9d)}
5a + 10d = 1/4 ( 5a + 35d)
20a + 40d = 5a + 35d
15a = -5d
15 ( 2 ) = -5d. ( given)
d = 30 / -5
d = -6
to show; a20 = -112
solving l.h.s
a + 19d = 2 + 19 ( -6 ) = 2 - 114 = -112 = r.h.s

Answer 2

Answer:

First Term = 2.

Let d be the common difference of the A.P.

$\boxed{Therefore,}$

the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d.

Sum of first five terms = 10 + 10d.

Sum of next five terms = 10 + 35d

According to the given condition,

=>10+10d =

1/4(10+35d)

=>40+40d =

10+35do

=>30 = −5d

=>d = −6

∴ a20 = a+(20−1)d = 2+(19)(−6)

= 2−114 = −112 .

Hence,

The 20th term of the ap is -112