Question

the sum of first n terms of an AP is 5n^2+3n if its mth term is 168, find the value of m also find the 11th term of an AP.​

Answer 1

Question:

The sum of first n terms of an AP is $\sf\:5n{}^{2}+3n$ its mth term is 168, find the value of m also find the 11th term of an AP.

Theory :

•Genral term of an AP

$\sf\:T_{n}=a+(n-1)d$

•For an AP

$\sf\:T_{n}=S_{n}-S_{n-1}$

•Common difference of An AP is given by

$\sf\:d=S_{2}-2S_{1}$

Solution :

Let $\sf\:a_{1},a_{2},a_{3}...a_{n}$ be the given AP.

Given: Sum of n terms $\sf=S_{n}=5n{}^{2}+3n$

$\implies\:\sf\:s_{n}=5n{}^{2}+3n...(1)$

Put n = 1

$\implies\:\sf\:S_{1}=5\times1{}^{2}+3\times1=8$

put n= 2

$\implies\:\sf\:S_{2}=26$

For an AP $\sf\:T_{n}=S_{n}-S_{n-1}$

First term,$\sf\:T_{1}=S_{1}-S_{1-1}$

$\implies\:\sf\:a_{1}=S_{1}-S_{0}$

$\sf\:a_{1}=5+3=8$

Second term,$\sf\:T_{2}=S_{2}-S_{2-1}$

$\sf\:a_{2}=S_{2}-S_{1}$

$\sf\:a_{2}=26-8=18$

Common difference, $\sf\:d=a_{2}-a_{1}$

$\sf\:d=18-8=10$

Given mth term is 168

$\sf\:T_{m}=a+(m-1)d$

$\sf168 = 8 + (n- 1) \times 10$

$160 = (m - 1) \times 10$

$\sf \frac{16 \times \cancel{10}}{ \cancel{10}} = m - 1$

$\sf m = 16 + 1 = 17$

Now the 11th term of an AP

$\sf\:T_{11}=a+(10-1)d$

$\implies\:\sf\:T_{11}=8+9d$

$\implies\:\sf\:T_{11}=8+9\times10$

$\implies\:\sf\:T_{n}=108$

Therefore, the value of m =17

and 11th term of an AP = 108

___________________________

More About Arithmetic progression:

The sum of first n terms of an AP is given by ;

$\sf S_{n} = \dfrac{n}{2} (2a + (n - 1)d)$

Answer 2

Question :

• the sum of first n terms of an AP is 5n^2+3n if its mth term is 168, find the value of m also find the 11th term of an AP.

Answer :

Sum of first n terms=$S_{n}$

=5n²+3n

Putting n=1, S₁=t₁=5+3=8

Putting n=2, S₂=52²+32=20+6=26

∴, t₂=S₂-S₁=26-8=18

Putting n=3, S³=53²+33=45+9=54

∴, t₃=S₃-S₂=54-26=28

∴, First term=a=8,

common difference=28-18=18-8=10

∴, the mth term=$t_{m}$

=8+(m-1)10=168

or, (m-1)10=168-8

or, m-1=160/10

or, m-1=16

or, m=16+1

or, m=17

∴, the 20th term =$t_{20}$

=8+(20-1)10

=8+19×10

=8+190

=198

∴, m=17 and $t_{20}$=198