An AP consists of 50 terms of which 3rd term is 12and the last term is 106.find the 29th term
Answers
Answered by
5
Answer:
Let a , d are first term and common
difference of an A.P
nth term = Last term = a + ( n - 1 )d
an = a + ( n - 1 )d
Now ,
It is given that ,
Third term = 12
a + 2d = 12 ------( 1 )
Last term = 106
a + 49d = 106 ---( 2 )
Subtract ( 1 ) from ( 2 ) , we get
47d = 94
d = 2
Substitute d value in equation ( 1 ) ,
We get
a + 2 × 2 = 12
a = 12 - 4
a = 8
29th term = a + 28d
a29 = 8 + 28 × 2
= 8 + 56
= 64
Answered by
2
Answer:
29n = 64
Step-by-step explanation:
Given -
n = 50, 3n = 12 , 50n, 106
To Find-
29n = ?
Solution
3n = 12 (given)
so,
a + (n -1)d = 12
a + (3-1)d = 12
a + 2d = 12 .....(1)
50n = 106 (given)
a + 49d = 106 .....(2)
Using Elmination Method
From Equation 1 and 2
See Pic
Then We Have
-47d = - -94
d = -94/-47
d = 2
Putting The Value Of d in Equation 1
a + 2× 2 = 12
a + 4= 12
a = 12-4
a = 8
So,
a¹ = 8
a² = a +d = 8+2= 10
a³ = a +2d =8+2×2 = 8+4 = 12
So The AP Is,
8, 10 , 12,……...………………….
NOW -
29n = a + 28d
= 8 + 28×2
= 8+ 56
= 64
So,
29n = 64
❣️ Hope This Help You ❣️
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