athang
Answered

Solve Q.IV)1 from the picture i.e. For an A.P.........................

Solve QIV1 from the picture ie For an AP class=

Answer :

kvnmurty
[tex]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\\ [/tex]


Mathexpert
Given a = 6 and S7 = 105
[tex]S_n = \frac{n}{2}(a+l) [/tex]
[tex]S_7 = \frac{7}{2}(6+l) [/tex]
[tex]\frac{7}{2}(6+l) = 105[/tex]
[tex]6 + l = 30 \Rightarrow l = 24[/tex]
That means, 7th term = 24
a + 6d = 24
6 + 6d = 24
1 + d = 4
d = 3

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

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


Now, from equations (1) and (2)

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

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