Question

for what values a are 3a+24,49,29and6a+31 sequentinal member of an aethemitic progression

Answer 1

Answer:

• The value of a is 5/3.

Given:

• 3a +24, 49, 29 and 6a + 31 are sequential members of an arithmetic progression.

To find:

• The value of a.

Solution:

Here,

$\sf{t_{1}}$ = 3a + 24,

$\sf{t_{2}}$ = 29,

$\sf{t_{3}}$ = 49,

$\sf{t_{4}}$ = 6a + 31

If they are in arithmetic progression, then

$\sf{t_{2}-t_{1}=t_{3}-t_{2}}$

$\sf{\therefore}$ 49 - (3a + 24) = 29 - 49

$\sf{\therefore}$25 - 3a = 20

$\sf{\therefore}$ -3a = -5

$\sf{\therefore}$ a =5/3

The value of a is 5/3.