Question

The 11th term and the 21st term of an AP are 16 and 29 respectively find the 41 St term of this AP

Answer 1

$the \: 11th \: term \: \\ = a + 10d = 16...........(i) \\ 21st \: term \: is \\ = a + 20d = 29.............(ii) \\ now \: substract \: eq(i)from(ii) \\ = > (a + 20d) - (a + 10d) = 29 - 16 \\ = > a + 20d - a - 10d = 13 \\ = > 10d = 13 \\ = > d = 1.3 \\ put \: this \: in \: equation \: (i) \: we \: get \\ = > a + 10 \times 1.3 = 16 \\ = > a + 13 = 16 \\ = > a = 16 - 13 = 3 \\ \\ now \: \\ common \: difference = 1.3 \\ first \: term = 3 \\ hence \: 41th \: term \\ = > a + (n - 1) \times d \\ > 3 + (41 - 1) \times 1.3 \\ = > 3 + 40\times 1.3 \\ = > 3 + 52 \\ = > 55$
Hence 41th term is 55.


I hope this helps....

Answer 2

11th term =a+10d=16 ...............(1)
21st term=a+20d=29..............(2)
equal 1 - equal 2
a+10d=16
(-) a+20d=29
_________
0 -10d = -13
d=13/10

put d =-13/10 in 1

a+10 x 13/10= 16
a +13 =16
a= 3

41 St term = a+40d
= 3 + 40 x 13/10==3 + 4 x13
= 3 +52= 55

hope this helps