the sum of first n terms of an ap whose 1st term is 8 and common difference is 20, is equal to the sum of first 2n term of another AP whose 1st term is -30 and common difference is 8 ,find n.
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the sum of first n terms of an ap whose 1st term is 8 and common difference is 20, is equal to the sum of first 2n term of another AP whose 1st term is -30 and common difference is 8 ,find n.
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GIVEN :–
The sum of first n terms of an A.P. whose 1st term is 8 and common difference is 20, is equal to the sum of first 2n term of another AP whose 1st term is -30 and common difference is 8.
TO FIND :–
• Value of 'n' = ?
SOLUTION :–
• If 'a' is first term and 'd' is Common difference then sum –
[tex] \\ \: \: \longrightarrow \: \: \large { \boxed{ \sf S = \dfrac{n}{2}[2a + (n - 1)d)]}} \\ [/tex]
• According to the first condition –
[tex] \\ \: \implies \: \sf S_1 = \dfrac{n}{2}[2(8)+ (n - 1)(20)] \\ [/tex]
[tex] \\ \: \implies \: \sf S_1 = \dfrac{n}{2}[16+ 20n - 20] \\ [/tex]
[tex] \\ \: \implies \: \sf S_1 = \dfrac{n}{2}[20n - 4] \\ [/tex]
[tex] \\ \: \implies \: \sf S_1 = n(10n - 2) \\ [/tex]
[tex] \\ \: \implies \: \sf S_1 = 10{n}^{2} - 2n \: \: \: - - - eq.(1) \\ [/tex]
• According to the second condition –
[tex] \\ \: \implies \: \sf S_2 = \dfrac{2n}{2}[2( - 30)+ (2n - 1)(8)] \\ [/tex]
[tex] \\ \: \implies \: \sf S_2 = n[ - 60+ 16n - 8] \\ [/tex]
[tex] \\ \: \implies \: \sf S_2 = n[16n - 68] \\ [/tex]
[tex] \\ \: \implies \: \sf S_2 = 16{n}^{2} - 68n \: \: \: - - - eq.(2) \\ [/tex]
• Now According to the question –
[tex] \\ \: \implies \: \sf S_1 = S_2 \\ [/tex]
[tex] \\ \: \implies \: \sf 10{n}^{2} - 2n = 16{n}^{2} - 68n \\ [/tex]
[tex] \\ \: \implies \: \sf 10{n}^{2} - 16{n}^{2} = 2n - 68n \\ [/tex]
[tex] \\ \: \implies \: \sf -6{n}^{2}= - 66n \\ [/tex]
[tex] \\ \: \implies \: \sf -6{n}^{2} + 66n = 0\\ [/tex]
[tex] \\ \: \implies \: \sf n(n - 11) = 0\\ [/tex]
[tex] \\ \: \implies \: \sf n = 0 \: , \: n = 11 \\ [/tex]
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