Question

Find four consecutive terms in an A.P. such that
their sum is - 54 and the sum of the first and the
third terms is -30.
[4 Marks]​

Answer 1

Answer:

The required four consecutive terms in an A.P. are - 18, - 15, - 12, - 9.

Step-by-step-explanation:

Let the four consecutive terms in an A.P be a, a + d, a + 2d, a + 3d.

From the first condition,

( a ) + ( a + d ) + ( a + 2d ) + ( a + 3d ) = - 54

⇒ a + a + d + a + 2d + a + 3d = - 54

⇒ 4a + 6d = - 54

By dividing both sides by 2, we get,

⇒ 2a + 3d = - 27 - - ( 1 )

From the second condition,

( a ) + ( a + 2d ) = - 30

⇒ a + a + 2d = - 30

⇒ 2a + 2d = - 30

By dividing both sides by 2, we get,

⇒ a + d = - 15

⇒ a = - 15 - d

⇒ a = - d - 15 - - ( 2 )

Now,

2a + 3d = - 27 - - ( 1 )

⇒ 2 ( - d - 15 ) + 3d = - 27 - - [ From ( 2 ) ]

⇒ - 2d - 30 + 3d = - 27

⇒ - 2d + 3d = - 27 + 30

⇒ d = 3

By substituting d = 3 in equation ( 2 ), we get,

a = - d - 15 - - ( 2 )

⇒ a = - 3 - 15

⇒ a = - 18

Now,

First term = a

⇒ a = - 18

Now,

Second term = a + d

⇒ a + d = - 18 + 3

⇒ a + d = - 15

Now,

Third term = a + 2d

⇒ a + 2d = - 18 + 2 ( 3 )

⇒ a + 2d = - 18 + 6

⇒ a + 2d = - 12

And,

Fourth term = a + 3d

⇒ a + 3d = - 18 + 3 ( 3 )

⇒ a + 3d = - 18 + 9

⇒ a + 3d = - 9

∴ The required four consecutive terms in an A.P. are - 18, - 15, - 12, - 9.

Answer 2

$\huge\red{Refer-Attachment}$

The four consecutive terms in A.P. are -18 , -15 , -12 , -9