Question

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

Answer 1

Correct Question →

The sum of first n terms of an AP is 5n² + 3n , if it's mth term is 168 , find the value of m also find the 11th term of AP .

Answer :-

$\boxed{\mathsf{11th\: term\: of \: AP \: = \: 198}}$

$\boxed{\mathsf{Value\:of\:m\:is\:17}}$

Solution :-

It's given that sum of n terms of AP is 5n² + 3n

Putting the value of n as 1

→ $S_1$ = 5( 1 )² + 3( 1 )

→ $S_1$ = 5 + 3 = 8

Now putting n as 2

→ $S_2$ = 5 (2)² + 3(2)

→ $S_1$ = 20 + 6 = 26

$S_1$ represent the first term of AP .

→ So a1 = 8

$S_2$ represents the sum of first two terms of AP.

And we have first term of AP so by subtracting first term from sum of two terms of AP we will get the second term of AP .

→ 26 - 8

→ a2 = 18 .

Now we have first two terms of the given AP .

$\boxed{\mathsf{Common \:difference\: = \:a1 \:-\: a2 }}$

→ d = 18 - 8

→ Common difference = 10 .

Now we are given that mth term of AP is 168.

→ 168 = a1 + ( m-1 ) d

→ 168 = 8 + ( m-1 ) 10

→ 168 - 8 = 10m - 10

→ 160 + 10 = 10m

→ $\frac{170}{7}$ = m

→ m = 17

So the value of m is 17 .

now we have to find the value of 11th term of the AP .

$\boxed{\mathsf{a_n = a_{1} + (n-1)d}}$

→ 11th term = a1 + 10 d

→ 8 + 10(10)

→ 8 + 100

→ 108

So the 11th term of AP is 108 .

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$\boxed{ \mathsf{s_n \: = \: \frac{n}{2} (2a + (n - 1)d)}}$

Answer 2

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

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More About Arithmetic progression:

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

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