#### 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

#### Solution

(a) Let 'a' be the first term and 'd' the common difference of the given A.P.

For t_{11}= 16, n ==11, we have

t_{11} = a(11-1)d

∴16 = a+10d ..........(1)

for t_{21}=29,n=21,we have

t_{21} = 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

∴t_{34}= 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