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AnantPawanChaturvedi
@AnantPawanChaturvedi
September 2023
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The third term of an a. p. is 7 and its 7th term is 2 more than thrice of its third term. the sum of its first 20 terms equals
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Mysteryboy01
Verified answer
[tex]a_{3} = 7 \\ \\ a_{7} =3 a_{3} + 2 \\ \\ a_{7} = 3(7) + 2 \\ \\ a_{7} = 21 + 2 \\ \\ a_{7} = 23[/tex]
[tex] = \frac{a + 2d = 7...........(1)}{a + 6d = 23.........(2)} \\ \\ - 4d = 16 \\ \\ d = \frac{16}{ - 4} \\ \\ d = - 4 \: \: ........(3)[/tex]
[tex]Substitute \: 3 \: in \: 1[/tex]
[tex] S_{n} = \frac{n}{2} [ (2a + (n - 1)d ] \\ \\ S_{20} = \frac{20}{2} [ 2( - 1) + (20 - 1)4 ] \\ \\ S_{20} = 10 [ 2( - 1) + (19)4 ] \\ \\ S_{20} = 10 [ 76 - 2 ] \\ \\ S_{20} = 10 [ 74 ] \\ \\ S_{20} = 740[/tex]
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Answers & Comments
Verified answer
[tex]a_{3} = 7 \\ \\ a_{7} =3 a_{3} + 2 \\ \\ a_{7} = 3(7) + 2 \\ \\ a_{7} = 21 + 2 \\ \\ a_{7} = 23[/tex]
[tex] = \frac{a + 2d = 7...........(1)}{a + 6d = 23.........(2)} \\ \\ - 4d = 16 \\ \\ d = \frac{16}{ - 4} \\ \\ d = - 4 \: \: ........(3)[/tex]
[tex]Substitute \: 3 \: in \: 1[/tex]
[tex] S_{n} = \frac{n}{2} [ (2a + (n - 1)d ] \\ \\ S_{20} = \frac{20}{2} [ 2( - 1) + (20 - 1)4 ] \\ \\ S_{20} = 10 [ 2( - 1) + (19)4 ] \\ \\ S_{20} = 10 [ 76 - 2 ] \\ \\ S_{20} = 10 [ 74 ] \\ \\ S_{20} = 740[/tex]