Question

if sum of the n terms of an Ap is 3n²+4n and n term is 151 then the value of n is​

Answer 1

Answer:

The value of n is 25.

Step-by-step explanation:

Given :

Sum of the n terms of an A.P is 3n² + 4n and nth term is 151.

To find :

the value of n

Solution :

Sum of the n terms of an A.P is 3n² + 4n

Put n = 1, the result we get is the first term.

= 3(1)² + 4(1)

= 3 + 4

= 7

∴ The first term, a = 7

• nth term of an A.P is given by,

     $\longmapsto \bf a_n=a+(n-1)d$

where

a is the first term

d is the common difference

nth term = 151

a + (n - 1)d = 151    

• Sum of n terms of an A.P is given by,

     $\longmapsto \bf S_n=\dfrac{n}{2}[2a+(n-1)d]$

$\sf \dfrac{n}{2}[2a+(n-1)d]=3n^2+4n \\\\ \sf \dfrac{n}{2}[a+a+(n-1)d]=3n^2+4n \\\\ \sf \dfrac{n}{2}[a+151]=3n^2+4n \ \ [\because a + (n - 1)d = 151] \\\\ n(a+151)=2(3n^2+4n) \\\\ \sf n(7+151)=6n^2+8n \ \ [\because a=7] \\\\ \sf 158n-8n=6n^2 \\\\ \sf 150n=6n^2 \\\\ \sf 150=6n \\\\ \sf n=150/6 \\\\ \longrightarrow \sf n=25$

Therefore, the value of n is 25