find the sum of first n terms of an ap is given by 2n square +3n.find the16 term of Ap
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Sn = 2n^2 + 3n
n/2 ( 2a + (n-1)d) = 2n^2 + 3n
n/2 ( 2a + nd - d ) = 2n^2 + 3n
2an/2 + n^2d/2 - nd/2 = 2n^2 + 3n
2an + n^2d - nd = 4n^2 + 6n
2an + n^2d - nd - 4n^2 - 6n = 0
n ( 2a + nd - d ) = (4n^2 + 6n)
2a + (n-1)(d) = 4n + 6
a + (n-1)(d) = 4n + 6 - a
an = 4n + 6 - a
a16 = 4(16) + 6 - a
a16 = 64 - 6 - a
a16 = 58 - a
n/2 ( 2a + (n-1)d) = 2n^2 + 3n
n/2 ( 2a + nd - d ) = 2n^2 + 3n
2an/2 + n^2d/2 - nd/2 = 2n^2 + 3n
2an + n^2d - nd = 4n^2 + 6n
2an + n^2d - nd - 4n^2 - 6n = 0
n ( 2a + nd - d ) = (4n^2 + 6n)
2a + (n-1)(d) = 4n + 6
a + (n-1)(d) = 4n + 6 - a
an = 4n + 6 - a
a16 = 4(16) + 6 - a
a16 = 64 - 6 - a
a16 = 58 - a
Ayushpratapsingh2210:
a = 2(1^2) + 3(1) = 5
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