Question

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

Answer 1

Answer:

73/4

Step-by-step explanation:

Given,

Sum of first 'n' terms of an A.P(S_n) = 4n² - 3n/4

To Find :-

Value of 10th term of an A.P

How To Do :-

Here as they given the value of sum of 'n' terms of an A.P we need to equate that value to the formula of sum of 'n' terms of an A.P . We can observe that the equation satisfies for any value of 'n' . By taking the value of 'n' as 1 we can find the value of 'a' and by taking the value of 'n' as 2 we need to find the value of 'd' and by using those we need to find the value of a_4.

Formula Required :-

Sum of 'n' terms of an A.P :-

S_n = n/2[2a + (n - 1)d

nth term of an A.P :-

a_n = a + (n - 1)d

Solution :-

S_n = 4n² - 3n/4

$\dfrac{n}{2}(2a + (n - 1)d) = \dfrac{4 {n}^{2} - 3n}{4}$

∴ The above equation satisfies for any value of 'n'.

[ S_n = n/2[2a + (n - 1)d ]

Substituting the value of 'n' as 1 :-

$\dfrac{1}{2}(2a + (1 - 1)d) = \dfrac{4( {1}^{2} ) - 3(1)}{4}$

$\dfrac{1}{2}(2a + (0)d) = \dfrac{4 - 3}{4}$

$\dfrac{1}{2} (2a + 0) = \dfrac{1}{4}$

$\dfrac{2a}{2} = \dfrac{1}{4}$

a = 1/4

∴ First term = a = 1/4

Substituting value of 'n' as 2 :-

$\dfrac{2}{2}(2a + (2 - 1)d) = \dfrac{4( {2}^{2} ) - 3(2)}{4}$

1(2(1/4) + (1)d) = (4(4) - 6)/4

1(1/2 + d) = (16 - 6)/4

1/2 + d = 10/4

1/2 + d = 5/2

d = 5/2 - 1/2

d = (5 - 1)/2

d = 4/2

d = 2

∴ common difference = d = 2.

10th term of an A.P (a_10) :-

a_10 = a + (10 - 1)d

Substituting the values of 'a' and 'd' :-

= 1/4 + (9)2

= 1/4 + 18

= (1 + 18(4))/4

= (1 + 72)/4

= 73/4

∴ 10th term of an A.P = 73/4