# 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.

Given, first term a = 2

Let d be the common difference

Let Sn denote sum of n terms,

tn denote nth term.

tn = a + (n – 1)d

S1 = Sum of first five terms.

S2 = Sum of next five terms.

S1 = 10 × (1 + d)

Sum of next 10 terms = Sum of first 10 terms - Sum of first 5 terms

S2 = 20 + 45d – 10 – 10d = 10 + 35d

Also, S1 = 1/4 S2

10 + 10d = 1/4 (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

t20 = 2 + (20 – 1) × (-6)

t20 = 2 – 19 × 6 = 2 – 114

t20 = -112

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