An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th
term.
Answers
Answer:
We know that the formula for the nth term is t
n
=a+(n−1)d, where a is the first term, d is the common difference.
It is given that third term of an A.P is t
3
=12, therefore,
t
n
=a+(n−1)d
⇒12=a+(3−1)d
⇒a+2d=12......(1)
Also, it is given that the 50th term is t
50
=106, therefore,
t
n
=a+(n−1)d
⇒106=a+(50−1)d
⇒a+49d=106......(2)
Now, subtract equation 1 from 2 as follows:
(a−a)+(49d−2d)=106−12
⇒47d=94
⇒d=
47
94
=2
Substitute the value of the difference d=2 in equation 1:
a+(2×2)=12
⇒a+4=12
⇒a=12−4=8
Now, the 29th term with a=8 and d=2 can be obtained as:
t
29
=8+(29−1)2=8+(28×2)=8+56=64
Hence, the 29th term is 64
Answer:
64
Step-by-step explanation:
Let the first term of the AP be 'a' and common difference be 'd'.
3rd term = 12 ⇒ a + 2d = 12 ...(1)
50th term = 106 ⇒ a + 49d = 106 ...(2)
Subtract (1) from (2), we get 47d = 94
⇒ d = 94/47
⇒ d = 2
Hence, in (1), a + 2(2) = 12 ⇒ a = 8
∴ 29th term = a + 28d
= 8 + 28(2)
= 64