If Sn denotes the sum of n terms of an AP and if S1= 6 , S7 = 105 , - Maths - Arithmetic Progressions

Question

If Sn denotes the sum of n terms of an AP and if S1= 6 , S7 =
105 , then show that Sn : S(n-3) = (n+3) : (n-3)

Aakash Sharma · Accepted Answer

Let the first term and common difference of the given A.P. be a and d respectively.

Then S₁ = a₁ = a = 6

$S_{7} = \frac{7}{2} [2 a + 6 d] = 105 \Rightarrow a + 3 d = \frac{105}{7} = 15 \Rightarrow 3 d = 15 - a = 15 - 6 = 9 \Rightarrow d = \frac{9}{3} = 3$

$Thus \frac{S_{n}}{S_{n - 3}} = \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{(n - 3)}{2} [2 a + ((n - 3) - 1) d]} = \frac{n [12 + 3 n - 3]}{(n - 3) [12 + 3 n - 12]} = \frac{3 n [n + 3]}{(n - 3) (3 n)} = \frac{(n + 3)}{(n - 3)}$

Hence S_n : S_n–₃= (n + 3) : (n – 3)

If Sn denotes the sum of n terms of an AP and if S1= 6 , S7 = 105 , then show that Sn : S(n-3) = (n+3) : (n-3)