Question

in an AP the sum of first n terms is 3n^2/2+5n/2 find its 25 th term​

Answer 1

AnswEr :

• 25th term of the AP is 76 .

Step-by-step explanation :

GivEn :

• Sum of the 1st n terms is Sn = 3n²/2 + 5n/2

To find :

• 25th term of an AP = ?

SoluTion :

As we're given with sum of n terms of an AP we know the formula that,

nth term of an AP = sum of nth term - sum of (n-1)th term

Substituting the values, [ n = 25 ]

$\sf :\implies (sum \: of \: 25th \: term) - (sum \: of \: (25-1)th \: term)$

$\sf :\implies (sum \: of \: 25th \: term) - (sum \: of \: 24th \: term)$

According to the given eqn,

$\sf :\implies \: \: \bigg(\dfrac{3 \times {25}^{2}}{2} + \dfrac{5 \times 25}{2} \bigg) - \bigg(\dfrac{3 \times {24}^{2}}{2} + \dfrac{5 \times 24}{2} \bigg)$

$\sf :\implies \: \: \dfrac{1875}{2} + \dfrac{125}{2} - \dfrac{1728}{2} + \dfrac{120}{2}$

$\sf :\implies \: \: \dfrac{2000}{2} - (864 + 60)$

$\sf :\implies \: \:1000 - 924$

$\sf :\implies \: \: { \purple{76}}$

Therefore, 25th term of the AP is 76 .

• Related formulae :

1. The general form of an AP is,

• a, a + d, a + 2d, a + 3d...

2. nth term of an AP is

• an = a + (n − 1) × d

3. Sum of n terms of an AP is

• S = n/2[2a + (n − 1) × d]

Answer 2

Step-by-step explanation:

It is given that sum of n terms of A.P. is $\sf \dfrac{3n^2}{2} + \dfrac{5n}{2}$

Let the sum of n terms is given by $\sf S_n$

So,

$:\implies \sf S_n = \Bigg \lgroup\dfrac{3n^2}{2} + \dfrac{5n}{2} \Bigg \rgroup$

Now, put n = 1 :

$:\implies \sf S_1 = \Bigg \lgroup\dfrac{3(1)^2}{2} + \dfrac{5(1)}{2} \Bigg \rgroup$

$:\implies \sf S_1 = \Bigg \lgroup\dfrac{3}{2} + \dfrac{5}{2} \Bigg \rgroup$

$:\implies \sf S_1 = 4$

We know, $\sf S_1 = a_1$ as $\sf S_1$ is a sum of first term only.

Therefore, the first term is a = 4.

Now, Put n = 2 :

$:\implies \sf S_2 = \Bigg \lgroup\dfrac{3(2)^2}{2} + \dfrac{5(2)}{2} \Bigg \rgroup$

$:\implies \sf S_2 = \Bigg \lgroup\dfrac{3(4)}{2} + \dfrac{5(2)}{2} \Bigg \rgroup$

$:\implies \sf S_2 = \Bigg \lgroup\dfrac{12}{2} + \dfrac{10}{2} \Bigg \rgroup$

$:\implies \sf S_2 =6 + 5$

$:\implies \sf S_2 = 11$

_______________________

As we know that,

$\bigstar\:\: \sf S_2 = a_1 + a_2\:\:\bigstar$

$\dashrightarrow\:\: \sf 11 = 4 + a_2$

$\dashrightarrow\:\: \sf a_2 = 11 - 4$

$\dashrightarrow\:\: \sf a_2 = 7$

As,

$\sf a_1 = 4$ and $\sf a_2 = 7$

So, common difference (d) will be :

$\bigstar \: \: \sf d = a_2 - a_1$

$: \implies \sf d = 7 - 4$

$: \implies \sf d = 3$

Thus,

$\bigstar\:\: \sf a_n = a + (n - 1) d\:\:\bigstar$

$\leadsto \sf a_{25} =4 + (25 - 1) \times 3$

$\leadsto \sf a_{25} =4 + 24\times 3$

$\leadsto \sf a_{25} =4 + 72$

$\leadsto \sf a_{25} =76$

⠀

$\therefore\:\underline{\textsf{The 25th term of the AP is \textbf{76}}}.$