The 11th term and 21st term of an A.P. are 16 and 29 respectively then find a. The first term and the common difference b. The 34th term c. ‘n’ such that tn = 55.

Answer:

11 = 16   and t21 =  29    [Given]

We know that,

tn =  a + (n – 1 )d
t11 = a + (11 – 1 )d
∴ 16 = a + 10d    ------ eq. no. 1

∴ t21 = 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 34th Term

t34 = a + (34 – 1 )d
t34 = 3 + 33 (1.3)
t34 = 3  + 42.9
t34 = 45.9

To find .   ‘n’ such that tn = 55

tn = 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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