|
Answer» Sum of 3 terms of AP is 27
so, a +(a+d)+(a+2d) = 27
3a + 3d = 27
3(a+d) = 27
a+d = 27/3
a+d = 9.......(a)
where a = 9 - d
d = 9-a
taking SQUARE on both sides in (a)
so, (a+d)² = 9²
a² + d² + 2AD = 81
a² + d² = 81 - 2ad..........(1)
sum of their squares is 293
so, a² + (a+d)² + (a+2d)² = 293
a² + (a+d)² + ((a+d)+d)² = 293
a² + 9² + (9+d)² = 293
a² + 81 + 81 + 18d + d² = 293
a² +162 + 18d + d² = 293
a² + 18d + d² = 293 - 162
a² + 18d + d² = 131
a² + d² = 131 - 18d........(2)
equate (1) and (2)
81 - 2ad = 131 - 18d
18d - 2ad = 131 - 81
2d(9 - a) = 50
2d(d) = 50
d² = 50/2
d² = 25
d = 5
a = 9 - 5 = 4
checking whether the ANSWER is correct or not:
substitute a and d value in
a + (a+d) + (a +2d) = 27
4 + (4+5) + (4+2*5) = 27
4+9+14 = 27
27 = 27
hence a and values are correct
so, the three values are 4, 9, 14
#mark as brainliest
|
