The 5th term of an A.P of $n$ terms whose sum is $n^{2} - 2 n$ is

5

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7

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8

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16

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Solution

The correct option is A $5$
$S_{n} = n^{2} - 2 n$

$\frac{n}{2} (2 a + (n - 1) d) = n (n - 2)$

$\frac{2 a + (n - 1) d}{2} = n - 2$

To get, $a_{5} = a + 4 d$

we put $n = 9$

$\frac{2 a + (9 - 1) d}{2} = 9 - 2$

$a + 4 d = 7 = 5^{t h} t e r m$

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The 5th term of an A.P of n terms whose sum is n2−2n is

The correct option is A 5Sn=n2−2n n2(2a+(n−1)d)=n(n−2)2a+(n−1)d2=n−2To get, a5=a+4dwe put n=92a+(9−1)d2=9−2a+4d=7=5thterm

The 5th term of an A.P of $n$ terms whose sum is $n^{2} - 2 n$ is

The correct option is A $5$
$S_{n} = n^{2} - 2 n$

$\frac{n}{2} (2 a + (n - 1) d) = n (n - 2)$

$\frac{2 a + (n - 1) d}{2} = n - 2$

To get, $a_{5} = a + 4 d$

we put $n = 9$

$\frac{2 a + (9 - 1) d}{2} = 9 - 2$

$a + 4 d = 7 = 5^{t h} t e r m$