Question

If in an A.P s1=6,s7=105,then sn:sn-3 is same as​

Answer 1

s1=1/2(2a) (a = first term)

S1= a = 6

a = 6

now, s7= 7/2(2a+(7-1)d)

s7 = 105 = 7/2(12+6d)

105x2/7 = 12+6d

15x2 = 12+6d

30-12 = 6d

18=6d

d=3

sn=n/2(2a+(n-1)d)

sn= n/2(12+(n-1)3) = n/2(12+3n-3)

sn= n/2(9+3n)....................(i)

Sn-3 = (n-3)/2 (2a+(n-3-1)d)

Sn-3 = (n-3)/2 (12+(n-4)3)

Sn-3 = (n-3)/2 (12+3n-12)

Sn-3 = (n-3)/2 (3n).............(ii)

(i):(ii)

Sn:Sn-3 = n/2(9+3n)÷[(n-3)/2 (3n)]

=> n(9+3n) ÷ 3n (n-3)

=> (9+3n)/3(n-3)

=>3(3+n)/3(n-3)

=> n+3:n-3 = Sn: Sn-3 for given values of question

Answer 2

Answer:

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