Question

The sum of fist m terms of an AP is ( 4m²  - m). If its nth term is 107, find the value of n. Also, Find the 21^st term of this AP. 

Answer 1

Let S_m denotes the sum of the first m terms of the AP. Then, 

S_m = 4m² -m

⇒ S_m-1 = 4 (m-1)² - (m-1) 

= 4(m² -2m +1) -(m-1)

= 4m² - 9m + 5 

Suppose a_m denote the m^th  term of the AP. 

∴ a_m = s_m - s_m-1 

= (4m² - m ) - ( 4m² - 9m +5) 

= 8m -5                  ..............(1)

Now,

a_n = 107                             (Given)

⇒ 8n - 5 = 107         [ From (1)]

⇒ 8n = 107 + 5=112

⇒ n= 14

Thus, the value of n is 14.
Putting m = 21 in (1), we get
a₂₁ = 8 × 21 - 5 = 168 - 5 =163
Hence, the 21^st term of the AP is 163. 

 

Answer 2

\[S_m = 4 m^2 - m\]
\[\text{ We know } \]
\[ a_m = S_m - S_{m - 1} \]
\[ \therefore a_m = 4 m^2 - m - 4 \left( m - 1 \right)^2 + \left( m - 1 \right)\]
\[ a_m = 8m - 5\]
\[\text{ Now } , \]
\[ a_n = 107\]
\[ \Rightarrow 8n - 5 = 107\]
\[ \Rightarrow 8n = 112\]
\[ \Rightarrow n = 14\]
\[ a_{21} = 8\left( 21 \right) - 5 = 163\]