Question

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

Answer 1

(a) Let 'a' be the first term and 'd' the common difference of the given A.P.
For t₁₁= 16, n ==11, we have

t₁₁ = a(11-1)d

∴16 = a+10d         ..........(1)

for t₂₁=29,n=21,we have

t₂₁ = a+(21-1)d

∴29= a+20d         ...........(2)

Subtracting (1) from (2),we get

13 =10d⇒d=1.3

Substituting d = 1.3 in (1), we get

16=a+10(1.3)

∴16=a+13⇒1=3

Thus, the first term is 3 and the common difference is 1.3.

(b) For 34^th term, n = 34, a = 3, d = 1.3

t_n = a(n-1)d

∴t₃₄= 3(34-1)(1.3)

      =3+33*1.3

      =3+42.9

      =45.9∴

Thus, the 34^th term is 45.9.

(c) t_n = 55, a = 3, d = 1.3

∴t_n= a(n-1)d

∴55= 3(n-1)(1.3)

∴55-3= (n-1)(1.3)

∴(n-1)((1.3)=52

∴n-1= `52/1.3`

∴n-1= 40

∴n= 40+1

∴n= 41