Question

if in an ap s1 is 6 , S7 is 105, prove that Sn: Sn-3::(n+3):(n-3)​

Answer 1

Answer:

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

Given, S

1

=a=6 and

S

7

=105⇒

2

7

[2a+(7−1)d]=105

⇒

2

7

[2×6+6d]=105

⇒6+3d=15

⇒d=3

Now, S

n

=

2

n

[2a+(n−1)d]=

2

n

[12+(n−1)3]=

2

3n

[n+3] ...(1)

S

n−3

=

2

n−3

[2a+((n−3)−1)d]=

2

n−3

[12+(n−4)3]=

2

n−3

[3n] ...(2)

From (1) and (2), we get

S

n−3

S

n

=

2

n−3

[3n]

2

3n

[n+3]

=

n−3

n+3

Therefore, S

n

:S

n−3

::(n+3):(n−3)