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2014-08-27T16:01:15+05:30

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a = 6 \ \ \ \ S_7 = 105 \\ \\ S_n = a n + \frac{d (n-1) n }{2} = \frac{2 a n + n^2 d - n d}{2} \\ \\ S_7 = \frac{2 *6 * 7 + 7^2 d - 7 d}{2} = 105 \\ \\ 84 +42 d = 2 * 105\\ \\ 42 d = 126\\ \\ d = 3 \\ \\ S_n : S_{n-3} = \frac{12 n + 3n^2 - 3 n}{2} : \frac{12 (n-3) + 3 (n-3)^2 - (n-3) 3}{2} \\ \\ 3n^2 + 9n : 12n - 36 + 3n^2 - 18 n + 27 - 3n + 9 = 3n^2 + 9 n : 3 n^2 -9 n \\ \\ n+3 : n - 3 \ \ \ \ after striking\ off\ n\ and\ 3\\


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You have considered S_7 as 7th term but it should be sum of first 7 terms.
2014-08-27T16:24:11+05:30
Given a = 6 and S7 = 105
S_n =  \frac{n}{2}(a+l)
S_7 =  \frac{7}{2}(6+l)
\frac{7}{2}(6+l)  = 105
6 + l = 30 \Rightarrow l = 24
That means, 7th term = 24
a + 6d = 24
6 + 6d = 24
1 + d = 4
d = 3

S_n =  \frac{n}{2}[2a + (n-1)d]
S_n =  \frac{n}{2}[12 + 3n - 3]
S_n =  \frac{3n}{2}[3 + n]     ......(1)

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


Now, from equations (1) and (2)

S_n : S_(n-3) = \frac{3n}{2}[3 + n] : \frac{(n-3)}{2}[3n]
S_n : S_(n-3) = (n+3) : (n-3)
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