Question

If the sum of n terms of an AP is 3n²+ 5n then which of its term is 164.​

Answer 1

Answer:

Let a1,a2,a3,...... be the 1st,2nd,3rd....... terms of the AP and d be the common difference .

Given that

$S_n = {3n}^{2} + 5n$

So,

$S_1 = 3( {1)}^{2} + 5 (1)\\ \\ = 3 + 5 \\ \\ = 8 \\ \\ S_2 = 3(2 {)}^{2} + 5(2) \\ \\ = 12 + 10 \\ \\ = 22$

$S_3 = 3(3 {)}^{2} + 5(3) \\ \\ = 27 + 15 \\ \\ = 42$

$S_4 = 3( {4)}^{2} + 5(4) \\ \\ = 48 + 20 \\ \\ =68$

Now

$a_1 = S_1 = 8$

$a_2 = S_2 - S_1 \\ \\ = 22 - 8 \\ \\ = 14$

$a_3 = S_3 - S_2 \\ \\ = 42 - 22 \\ \\ = 20$

So,

Common difference of the AP is

d = 6

Now

Let 164 be the nth term of the AP .

So,

a = 8

d = 6

$a_n = 164$

n = ?

As

$a_n = a + (n - 1)d \\ \\ 164 = 8 + (n - 1)6 \\ \\ 164 = 8 + 6n - 6 \\ \\ 164 = 6n + 2 \\ \\ 6n = 162 \\ \\ n = \dfrac{ \cancel{162}}{ \cancel{6 \: \:} } \\ \\ \blue{\Large{ \boxed {\boxed{\boxed{\boxed{n = 27}}}}}}$

So,

27th term of the given AP is equals to 164.

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Given :

the sum of n terms of an AP is 3n²+ 5n

an=164

To Find :

the nth term

ㅤㅤㅤㅤ

Let the first term of an ap be $\bold{'a'}$

second term be $\bold{a_{2}}$

and 'D' be the common difference between 1st term and 2nd term.

$\bold{\boxed{D=a_{2}-a}}$

here ,value of $a_{2}$ and a is not given so we need to find it.For this we have to find S1 and S2 first.

S1 and S2 can be find by substituting value in the given equation.

$S(1) = 3 {(1)}^{2} + 5(1)$

$\bold{\boxed{S(1) = 8}}$

$S(2) = 3 {(2)}^{2} + 5(2)$

$\bold{\boxed{S(2 )= 22}}$

Note :S1=a and $\bold{\boxed{D=a_{2}-a}}$=S2-S1

$\bold{\boxed{D=a_{2}-a}}$=22-8=14

a=8 and $\bold{\boxed{D=a_{2}-a}}$=14

Let's finding common difference

$D= a_{2}- a1 = 14 - 8 = 6$

We an formula here :

$\bold{\boxed{an = a + (n - 1)d}}$

$164 = 8 + (n - 1)6$

$= > 164 = 8 + 6n - 6$

$= > 164 = 2 + 6n$

$= > 162 = 6n$

$= > n = \frac{\cancel{{162}}}{\cancel{{6} }} = 27$

Therefore,$\bold{nth\: term \:is \:{\red{27}} }$

∴27th term of an ap is 164