Q. 4 B5.0( 2 Votes )

# Find A.P. if Tn,

Formula Used.

an = a + (n–1)d

a2 = a + (2–1)d

a2 = a + d

If 2nd term of A.P is given as 1

Then,

a + d = 1

we get a = 1–d ......eq 1

a12 = a + (12–1)d

a12 = a + 11d

If 12th term of A.P is given as –9

Then,

a + 11d = –9

we get a = –9–11d ......eq 2

Equating both eq 1 and eq 2

We get ;

1–d = –9–11d

11d–d = –9–1

10d = –10

d = = –1

Putting d in eq 1 we get ;

a = 1–(–1)

= 2

As a = 2 and d = –1

Then;

a1 = a + (n–1)d = 2 + (1–1)(–1) = 2

a2 = a + (n–1)d = 2 + (2–1)(–1) = 2 + (–1)×1 = 2–1 = 1

a3 = a + (n–1)d = 2 + (3–1)(–1) = 2 + (–1)×2 = 2–2 = 0

a4 = a + (n–1)d = 2 + (4–1)(–1) = 2 + (–1)×3 = 2–3 = –1

an = a + (n–1)d = 2 + (n–1)(–1) = 2 + 1–n = 3–n

The A.P is 2, 1, 0, –1……, 3–n

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