Question

The sum of first n terms of an A.P. is $3n^{2}$ + 4n. Find the 25th term of this A.P.

Answer 1

Answer:

The 25th term is 151.

Step-by-step explanation:

Given :  

Sn = 3n² +  4n  …………..(1)

Now , on Replacing n by (n –1) in eq (i),  

S(n – 1) = 3(n – 1)² + 4(n – 1)

nth term of the A.P., an = Sn – S(n – 1)

an = (3n² + 4n) – [3(n – 1)² + 4(n – 1)]

⇒ an =(3n² + 4n) – [3(n² + 1 - 2n) + 4n - 4]

[(a + b)² = a² + b² - 2ab]

⇒ an =(3n² + 4n) – [3n² + 3 - 6n + 4n - 4]

⇒ an = (3n² + 4n) – [3n² - 2n - 1]

⇒ an = 3n² + 4n – 3n² + 2n + 1  

⇒ an =  6n  + 1

For 25th term :  

a25 = 6 × 25 + 1

a25 = 150  + 1

a25 = 151

Hence, the 25th term is 151.

THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

151

Step-by-step explanation:

Sn=3n^2+4n

S1=3(1)^2+4(1)=7

Like this, S24=3(24)^2+4(24)

=3×576+96=1824

S25=3(25)^2+4(25)

=3(625)+100

=1875+100=1975

S25-S24=a25

1975-1824=a25

a25=151