Question

Find first 2 terms in AP for which 5n=n^2(n+c)

Answer 1

Answer:

First two terms are

$(1) a_{1} = 6 - 2(1) = 6 - 2 = 4\\\\ (2) a_{2} = 6 - 2(2) = 6 - 4 = 2$

Step-by-step explanation:

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

Now,

$S_{n}$ = 5n - n²

∴ By substituting n - 1 in place of the n 

We get,

$S_{n-1}$ = 5(n-1) - ( n - 1 )²

= 5n - 5 - ( n² + 1 - 2n )

= 5n - 5 - n² - 1 + 2n

= -n² + 7n - 6

∴$S_{n-1}$ = -n² + 7n - 6

Now, we know that

$n^{th} term = S_{n} - S_{n - 1}$

By substituting the values here, for $S_{n} and S_{n-1}$ we will get the value for term $n^{th}$.

$n^{th}$ term = 5n - n² - ( -n² + 7n - 6 )

$n^{th}$ term = 5n - n² + n² - 7n + 6

$n^{th}$ term = 6 - 2n

∴ The nth term of the given A.P is 6 - 2n.

First two terms are

$(1) a_{1} = 6 - 2(1) = 6 - 2 = 4\\\\ (2) a_{2} = 6 - 2(2) = 6 - 4 = 2$