If a,3a-1 and 3a+1 are three consecutive terms of an A.P ,then the value of a is
A) 2
B) 3/2
C) -3/2
D) -1
Answers
Answered by
1
Answer:
3/2 , B is correct option
Step-by-step explanation:
Given---> a , 3a - 1 , and 3a + 1 are consecutive terms of AP.
To find---> We know that if p , q and r are three consecutive term of AP then
q - p = r - q
=> q + q = p + r
=> 2q = p + r
Now, ATQ,
a , ( 3a - 1 ) , (3a + 1 ) are in AP , so
a + ( 3a + 1 ) = 2 ( 3a - 1 )
=> a + 3a + 1 = 2 × 3a - 2 × 1
=> 4a + 1 = 6a - 2
=> 4a - 6a = - 2 - 1
=> - 2a = - 3
=> 2a = 3
=> a = 3 / 2
Additional information---->
1) Formula for nth term of AP is
aₙ = a + ( n -1 )d
2) Formula for sum of n terms of AP is
Sₙ = n/2 { 2a + ( n - 1 ) d }
Answered by
6
option B= 3/2
#answerwithquality #bal
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