Question

In an A.P

$\sf S_n=4n^2+3n$
Find $\tt t_5$​

Answer 1

Answer:

Sn = 4n² + 3n

=> S (n + 1) = 4 (n + 1)² + 3 (n + 1)

S (n + 1) = 4n² + 8n + 4 + 3n + 3

S (n + 1) = 4n² + 3n + 8n + 7

Now, t (n + 1) = S (n + 1) - Sn

» t (n + 1) = 8n + 7

» t5 = 39

Answer 2

Given :-

Sn = 4n^2 + 3n

Solution :-

Let the n be 1

In AP, Sn = 4n^2 + 3n

Subsitute 1 In Sn

S1 = 4 * 1^2 + 3 * 1

S1 = 4 * 1 + 3

S1 = 4 + 3

S1 = 7

Therefore,

The sum of first term of an AP is 7

As we know that,

The sum of first term of an AP is first term

Therefore

t1 = 7

Now,

Let's take the n = 2 , Sn

Therefore,

S2 = 4 * (2)^2 + 3 * 2

S2 = 4 * 4 + 6

S2 = 16 + 6

S2 = 22

Thus,

Sum of second term of an AP is 22

Now,

Finding second term of an AP that is

tn = ?

First term + Second term = 22

Put the required values,

7 + t2 = 22

t2 = 22 - 7

t2 = 15

Thus, The second term of an AP is 15

t2 = a + d = 15

15 = 7 + d

d = 15 - 7

d = 8

Here ,

t = 7 and d = 8

Now, we have to find fifth term of an AP that is

t5 = ?

t5 = a + 4d

Put the required values,

t5 = 7 + 4 * 8

t5 = 7 + 32

t5 = 39

Hence, The fifth term of an AP is 39 $.$