Answer:

t_{11}
= 16 and t_{21} = 29
[Given]

We
know that,

t_{n}
= a + (n – 1 )d

∴ t_{11} = a +
(11 – 1 )d

∴ 16 = a + 10d  eq. no. 1

∴ t_{21} = a + (21 – 1) d

∴ 29 = a + 20 d eq. no. 2

Subtracting eq. 1 from eq. 2

29 = a + 20 d

16 = a + 10d





13 = 0 + 10d

∴ 13 = 10d

∴ 13/10 = d

∴ 1.3 = d

Substituting d = 1.3 in equation 1

∴ 16 = a + 10d

∴ 16 = a + 10(1.3)

∴ 16 = a + 13

∴ 16 – 13 = a

∴ 3 = a

To find the 34^{th} Term

t_{34} = a + (34 – 1 )d

t_{34} = 3 + 33 (1.3)

t_{34 }= 3 + 42.9

t_{34 }= 45.9

To find . ‘n’ such
that t_{n} = 55

t_{n} = a + (n – 1) d

∴ 55 = 3 + (n – 1) 1.3

∴ 55 – 3 = 1.3 (n – 1)

∴ 52 = 1.3 (n – 1)

∴ 52/1.3 = (n – 1)

∴ 40 = n – 1

∴ 40 + 1 = n

∴ n = 41

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