Question

if the sum to the first n terms an A.P is 4n²-3n the 10th term is​

Answer 1

Answer:Given Sum of n terms is S  

n

​

=3n  

2

−4n

So Sum of n−1 terms is S  

n−1

​

=3(n−1)  

2

−4(n−1)

=3n  

2

−6n+3−4n+4

=3n  

2

−10n+7

Sum of n terms is equal to sum of n−1 terms plus n  

th

 term

⟹n  

th

 term =S  

n

​

−S  

n−1

​

=3n  

2

−4n−(3n  

2

−10n+7)

n  

th

 term=6n−7

Step-by-step explanation:

Answer 2

Answer:

$a_{10} = 73$

Step-by-step explanation:

$S_n = 4 {n}^{2} - 3n$

Putting n = 1

$S_1 = 4 {(1)}^{2} - 3(1)$

$\implies \: 4 - 3$

$\implies \: \boxed{ \color{orange}1 = \: a_1}\: \: \: \: \dashleftarrow(1)$

Putting n = 2

$S_2 = 4 {(2)}^{2} - 3(2)$

$\implies \: 16 - 6$

$\implies \: \boxed{ \color{brown}10 = a_1 + a_2 }\: \: \: \: \: \dashleftarrow(2)$

We got the sum of 2 terms of an AP.

$a_1 + a_2 = 10$

$\implies \: a_1 + a_2 = 10$

$\implies \: 1 + a_2 = 10$

$\implies \: \color{purple}\boxed{a_2 = 9}$

Now common difference,

$d = a_2 - a_1$

$\implies \: 9 - 1$

$\implies \boxed{ \color{green}d = 8} \: \: \: \: \: \dashleftarrow(3)$

We know that,

$a_n = a + (n - 1)d$

$a_{10} = 1 + (10 - 1)8$

$a_{10} = 1 + 72$

$\color{red} \boxed{a_{10} = 73}$

Hence it's 10th term is 73