Question

the sum of the first three terms of an AP is 78and sum of their squares is 2190.find the number​

Answer 1

Answer:

17,26,35 ( when a=26 and d=9 )

35,26,17 ( when a=26 and d= -9 )

Step-by-step explanation:

given that sum of first three terms of an A.P is 78

=> (a-d) + a + (a+d) = 78

=> 3a = 78

=> a = 78/3

=> a = 26

also given the sum of the squares of the 3 terms is 2190

=> (a-d)²+ (a²)+ (a+d)² = 2190

=> a²-2ad+d²+a²+a²+2ad+d² = 2190

=> 3a²+ 2d² = 2190

=> 3 (26)² + 2d² = 2190

=> 3*676 + 2d² = 2190

=> 2028 + 2d² = 2190

=> 2d² = 2190-2028

=> 2d² = 162

=> d² = 162/2 = 81

=> d = ±9

taking d = 9 and a =26

the first 3 terms are (26-9), 26, (26+9)

=> 17,26,35

taking d = -9 and a =26

the first 3 terms are (26-(-9)), 26, (26+(-9))

=> 26+9 , 26 , 26-9

=> 35,26,17