Question

The sum of first m terms of an AP is ( 4m^2 - m ).If its nth term is 107, find the value of n. Also, find the 21st term of this AP.

Answer 1

Given the Sum of 1st 'm' terms of the AP as : 4m² - m

⇒ The 1st term of the Given AP can be found by substituting : m = 1

⇒ 1st term of the Given AP = 4(1²) - 1 = 3

⇒ a = 3

The Sum of First Two terms of the Given AP can be found by substituting : m = 2

⇒ The Sum of First two terms = 4(2²) - 2 = 16 - 2 = 14

⇒ The Second term of the Given AP can be found by subtracting the First term from the Sum of First two terms.

⇒ The Second term of the Given AP : 14 - 3 = 11

⇒ The Common Difference of the Given AP = 2nd term - 1st term

⇒  The Common Difference of the Given AP (d) = 11 - 3 = 8

Given that : The nth term is 107

⇒ a + (n - 1)d = 107

we got a = 3 and d = 8

substituting we get :

⇒ 3 + (n - 1)8 = 107

⇒ 3 + 8n - 8 = 107

⇒ n = 112/8 = 14

21st Term of the Given AP can be found by nth term formula :

⇒ nth term = a + (n - 1)d

⇒ 21st term = a + (21 - 1)d

⇒ 21st term = 3 + 20 × 8 = 163

Answer 2

Hey there !!


➡ Given :-

→ S $\tiny m$ = ( 4m² - m ).

→ nth term ( a $\tiny n$ ) = 107.


➡ To find :-

→ Value of n.

→ 21st ( a $\tiny 21$ ) term of AP.


➡ Solution :-

We have,

→ S $\tiny m$ = ( 4m² - m ).

Then,

→ S $\tiny m - 1$ = 4( m - 1 )² - ( m - 1 ) .

=> S $\tiny m - 1$ = 4 ( m² + 1 - 2m ) - m + 1.

=> S $\tiny m - 1$ = 4m² + 4 - 8m - m + 1.

=> S $\tiny m - 1$ = 4m² - 9m + 5.


▶ We know that :-

→ a $\tiny m$ = ( S $\tiny m$ - S $\tiny m - 1$ ).

=> a $\tiny m$ = ( 4m² - m ) - ( 4m² - 9m + 5 ).

=> a $\tiny m$ = 4m² - m - 4m² + 9m - 5.

=> a $\tiny m$ = 8m - 5.


▶ Now,

=> a $\tiny n$ = ( 8n - 5 ).

=> 107 = 8n - 5.

=> 8n = 107 + 5.

=> n = $\frac{112}{8}$

$\huge \boxed{ \boxed{ \bf => n = 14. }}$


▶Then,

→ a $\tiny 21$ = 8 × 21 - 5.

=> a $\tiny n$ = 168 - 5.

$\huge \boxed{ \boxed{ \bf => a \tiny n \large = 163. }}$

✔✔ Hence, it is solved ✅✅.

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