In an Ap whose first term is 2 ,the sum of first five terms is one fourth the sum of the next five terms.Show that T20=-112 find S20
Answers
Answer:
It is given that the first term of an A.P is a=2. Also, the sum of first five terms is one fourth the sum of the next five terms, therefore,
S
5
=
4
1
(S
10
−S
5
)
⇒4S
5
=S
10
−S
5
⇒4S
5
+S
5
=S
10
⇒5S
5
=S
10
We know that sum S
n
of n terms of an A.P with first term a and common difference d is:
S
n
=
2
n
[2a+(n−1)d]
Thus,
S
n
=
2
n
[2a+(n−1)d]
⇒5×
2
5
[2a+(5−1)d]=
2
10
[2a+(10−1)d]
⇒5[(2×2)+4d]=
5
2
×
2
10
[(2×2)+9d]
⇒5(4+4d)=2(4+9d)
⇒20+20d=8+18d
⇒20d−18d=8−20
⇒2d=−12
⇒d=−
2
12
⇒d=−6
We also know that the nth term of an A.P with first term a and common difference d is t
n
=a+(n−1)d, therefore, with a=2, n=20 and d=−6, we have
t
20
=2+(20−1)(−6)=2+(19×−6)=2−114=−112
Now,
S
20
=
2
20
[(2×2)+(20−1)(−6)]=10[4+(19×−6)]=10(4−114)=10×−110=−1100
Hence, T
20
=−112 and S
20
=−1100