Question

Q.5 Solve any one of the following.
(4)
1) The sum of five consecutive terms of an A. P. is 45. Sum of cubes of second
and fourth terms is 1944. Find the terms.
(Assume these terms as a - 2d, a-d, d, a + d, a + 2d)​

Answer 1

Answer:

3, 6, 9, 12, 15

Step-by-step explanation:

let the terms be

a - 2d, a - d, a, a + d, a + 2d

sum of the terms = 45

a - 2d + a - d + a + a + d + a + 2d = 45

5a = 45

a = 45/5 = 9

cube of 2nd term

${(a - d)}^{3} = {a}^{3} - {d}^{3} - 3 {a}^{2} d + 3a {d}^{2} \\ = {9}^{3} - {d}^{3} - 3( {9)}^{2} d + 3(9) {d}^{2} \\ = 729 - {d}^{3} - 243d + 27 {d}^{2}$

cube of 4th term

${(a + d)}^{3} = {a}^{3} + {d}^{3} + 3 {a}^{2} d + 3a {d}^{2} \\ = {9}^{3} + {d}^{3} + 3(81)d + 27 {d}^{2} \\ = 729 + {d}^{3} + 243d + 27 {d}^{2}$

sum of 2nd and 4th term = 1944

$729 - {d}^{3} - 243d + 27 {d}^{2} + 729 + {d}^{3} + 243d + 27 {d}^{2} = 1944 \\ 729 + 27 {d}^{2} + 729 + 27 {d}^{2} = 1944 \\1458 + 54 {d}^{2} = 1944 \\ 54 {d}^{2} = 1944 - 1458 \\ 54 {d}^{2} = 486 \\ {d}^{2} = \frac{486}{54} \\ {d}^{2} = 9 \\ d = \sqrt{9} = 3$

a = 9 d = 3

therefore the terms are

a - 2d = 9 - 2(3) = 9 - 6 = 3

a - d = 9 - 3 = 6

a = 9

a + d = 9 + 3 = 12

a + 2d = 9 + 2(3) = 9 + 6 = 15

3, 6, 9, 12, 15

hope you get your answer