The sum of n terms of an $A P$ is $5 n^{2} - 3 n$ . Find the AP and also its $10^{t h}$ term.

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Solution

Let $n = 1, 5 n^{2} - 3 n$
$= 5 - 3 = 2$
Putting $n = 2$ , $5 n^{2} - 3 n$
$= 14$
Therefore,
$2 n d$
term is $14 - 2 = 12$
Therefore $A . P$ is $2, 12, 22.....$ and so on.
$10 t h$ term $= a + 9 d$
$= 2 + 9 \times 10$
$= 2 + 90$
$= 92$

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The sum of n terms of an AP is 5n2−3n. Find the AP and also its 10th term.

Let n=1,5n2−3n=5−3=2Putting n=2,5n2−3n=14Therefore,2ndterm is 14−2=12Therefore A.P is 2,12,22..... and so on.10th term=a+9d=2+9×10=2+90=92

The sum of n terms of an $A P$ is $5 n^{2} - 3 n$ . Find the AP and also its $10^{t h}$ term.

Let $n = 1, 5 n^{2} - 3 n$
$= 5 - 3 = 2$
Putting $n = 2$ , $5 n^{2} - 3 n$
$= 14$
Therefore,
$2 n d$
term is $14 - 2 = 12$
Therefore $A . P$ is $2, 12, 22.....$ and so on.
$10 t h$ term $= a + 9 d$
$= 2 + 9 \times 10$
$= 2 + 90$
$= 92$