The 11th term and the 21st term of an A.P. are 16 and 29 respectively, find : (i) the 1st term and the common difference. (ii) the 34th term (iii) ‘n’ such that tn = 55.

8. The 11th term and the 21st term of an A.P. are 16 and 29 respectively, find :

(i) the 1st term and the common difference.

(ii) the 34th term

(iii) ‘n’ such that tn = 55.

Sol. tn = a + (n – 1) d

∴ t11 = a + (11 – 1) d

16 = a + 10d

∴  a + 10d = 16

∴ a = 16 – 10d .......(i)

∴ t21 = a + (21 – 1) d

29 = a + 20d

∴  a + 20d = 29 ......(ii)

Substituting eq. (i) in eq. (ii)

∴  a + 20d = 29 ......(ii)

∴ 16 – 10d + 20d = 29

∴ 10d = 29 – 16

∴ 10d = 13

∴ d = ¹³/₁₀

∴ d = 1.3

Substituting d = 1.3 in eq. (i)

a = 16 – 10d .......(i)

∴ a = 16 – 10 (1.3)

∴ a = 16 – 13

∴ a = 3

The first term is 3 and the common difference is 1.3

(ii) the 34th term

Sol. tn = a + (n – 1) d

t34 = a + (34 – 1) d

∴ t34 = 3 + 33 (1.3)

∴ t34= 3 + 42.9

∴ t34 = 45.9

∴ Thirty fourth term of A.P. is 45.9.

(iii) ‘n’ such that tn = 55. (1 mark)

Sol. tn = a + (n – 1) d

∴  55 = 3 + (n – 1) d

∴ 55 = 3 + (n – 1) 1.3

∴ 55 = 3 + 1.3n – 1.3

∴ 55 = 1.7 + 1.3n

∴ 53.3 = 1.3n

∴ n = ^53.3/_1.3

∴ n = ^53.3/_1.3 × ¹⁰/₁₀

∴ n = ⁵³³/₁₃

∴ n = 41