Math, asked by gudibandlasainathred, 9 months ago

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 rishu6845

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 Aɾꜱɦ

$<font color =“orange”>$

${ \huge \bf{ \mid{ \overline{ \underline{Answer}}} \mid}}$

$<body bgcolor= “purple” ><fontcolor=“white”>$

#answerwithquality #bal

$<marquee scrollamount=500>follow me </marquee>$

Previous Question

Next Question