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# 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 20^{th} term is - 112

**Solution:**

It is given that the first term of the A.P. a = 2

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

Hence, the A.P is 2, 2 + d, 2 + 2d, 2 + 3d...

We know that, S_{n} = n/2 [2a + (n - 1) d]

= 5/2 [2 x 2 + (5 - 1) d]

= 5/2 [4 + 4d]

= 10 + 10d

Sum of the first ten terms,

S_{10} = 10/2 [2a + (10 - 1) d]

= 5[4 + 9d]

= 20 + 45d

Hence,

Sum of the next five terms

S_{10} - S_{5} = (20 + 45d) - (10 + 10d)

= 20 + 45d - 10 - 10d

= 10 + 35d

According to the given condition,

⇒ 10 + 10d = 1/4 (10 + 35d)

⇒ 40 + 40d = 10 + 35d

⇒ 40d - 35d = 10 - 40

⇒ 5d = - 30

⇒ d = - 6

Therefore,

a_{20} = a + (20 - 1) d

= 2 + (19)(- 6)

= 2 - 114

= - 112

Thus, the 20^{th} term of the A.P is - 112

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 3

## 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 20^{th} term is - 112

**Summary:**

Given the above terms and knowing that the sum of A.P as S_{n} = n/2 [2a + (n - 1) d] we confirmed that the 20th term of the A.P is -112

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