Question

The sum of three consecutive terms in an A.P is 9 and the sum of their squares is 35. Find Sn. Correct answer is n2 or n(6-n)​

Answer 1

let ,three consecutive terms in a AP is a -d , a , a+ d
From 1st condition
a-d+a+a+d = 9
3a =9
a = 9/3
a =3

From 2nd condition
(a-d)^2 +a^2 + (a+d)^2 =35
a^2 -2ad+d^2 +a^2 + a^2 +2ad +d^2= 35
3^2 +3^2+ 3^2 + 2d^2 =35
9+9+9 +2d^2 =35
27 +2d^2 =35
2d^2=35-27
2d^2= 7
d^2 = 7/2
by taking square root
d = underoot 7/2

d = 1.87

Sn = n/2 [ 2a + (n-1) ×d]
Sn = n/2 [ 2×3 +(n-1) × 1.87 ]
Sn = n/2 [ 6 + (n-1) ×1.87 ]
Sn = n×3