Question

Find four consecutive terms in an A.P. whose sum is 36 and the product of the 2nd
and the 4th is 105. The terms are in ascending order.

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Answer 1

Answer:

this is your answer to the question

Answer 2

Given :

Sum of four consecutive terms in an A.P. = 36.

Product of second and Fourth terms in an A.p. = 105.

Terms are in ascending order

To find :

Four consecutive terms in an A.P.

Solution :

Let the four consecutive terms in an A.P are

( a- 3d), (a-d), (a+d), (a+3d)

According to the first condition

(a-3d) + (a-d) + (a+3d) + (a+d) = 36 → (1)

arranging' a' and 'd' together.

(a + a + a +a ) + ( -3d -d + 3d + d) =36 → (2)

4a + 0 = 36

4a = 36

a = 36/4

a= 9 → (3)

According to the second condition.

(a - d ) ( a +3d ) = 105

a ( a + 3d ) - d ( a + 3d ) = 105

a² + 3ad - ad - 3d² = 105

a² - 2ad - 3d² =105 → (4)

substituting value of a in equation (4)

(9)² -2 ( 9) d -3d² =105

81 -18d - 3d² = 105

- 3d² -18d + 81 -105 = 0

-3d -18d + 24 = 0

multiplying with (-1)

3d² + 18d - 24 =0

Dividing by 3

d² + 6d -8 = 0

Factorising the equation.

d² + 4d -2d -8 =0

d (d +4) -2 (d +4) =0

(d -2) ( d +4) = 0

d = 2, d = 4

( a- 3d), (a-d), (a+d), (a+3d)

{ 9-(3x 2 ) } , ( 9-2) ,(9+2) (9 +6)

( 3, 7, 11, 15 )

taking d =4

(9 - 12) , ( 9- 4), ( 9+4) , (9 + 12)

-3, 5, 13, 21

Hence the terms in an A.P. are

(3,7,11,15)

or

(-3,5,13,21)