Posted by **akash** on Tuesday, February 28, 2012 at 3:32am.

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**, Tuesday, February 28, 2012 at 11:24am
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

- arithmetic -
**Sebin PS**, Monday, March 5, 2012 at 3:58am
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

- Trigonometric Function -
**7.1**, Wednesday, December 7, 2016 at 6:20am
2ab/a² b²

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