# In an A.P., 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.

a = 2, S5 = (1/4) (S10 - S5)

S5= (5/2)(2 2 + 4d) = (5/2)(4 + 4d) = 10 + 10d

S10 = (10/2)(2 2 + 9d) = (10/2)(4 + 9d) = 5(4 + 9d) = 20 + 45d

Since S5 = (1/4) (S10 - S5)

10 + 10d = (1/4)(20 + 45d – 10 - 10d)

40 + 40d = 10 + 35d

-5d = 30

d = -6

a20 = 2 + (20 –1)(-6)

= 2 + (20 –1)(-6)

= 2 – 114

= - 112

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