Question

Find 3 Numbers in AP whose Sum is 12 and Sum of Cubes is 408

Answer 1

Let the numbers be (a - d), a, (a + d)

According to first condition,

a - d + a + a +d = 12

=> 3a = 12

=> a = 4

According to second condition,

(a - d)^3 + a^3 + (a+d)^3 = 408

=> ( 4 - d)^3 + (4)^3 + (4 +d)^3 = 408

=> 64 - d^3 - 12d(4 - d) + 64 + 64 + d^3 + 12d( 4 + d) = 408

=> 3 × 64 - 48d + 12d^2 + 48d + 12d^2 = 408

=> 192 + 24d^2 = 408

=> 24 d^2 = 216

=> d^2 = 9

=> d = 3

First number = ( 4- 3) = 1

Second number = 4

Third number = ( 4 + 3) = 7

Answer 2

Let three no in AP be α-β , α , α+β

ATQ,

α-β+α+α+β=12

3α=12

α=4

again ATQ

(α-β)³+α³+(α+β)³=408

α³-3α²β+3αβ²-β³+α³+α³+3α²β+3αβ²+β³=408

3α³+6αβ²=408

3(α³+2αβ²)=408

α³+2αβ²=136

putting the value of α

4³+2×4β²=136

64+8β²=136

8β²=136-64

8β²=72

β²=9

β=√9=3

3 no in AP are 1,4,7

Hope this help!

plzzzz mark it as brainlist.............