Question

The sum of first m terms of an AP is (4m² – m). If its nth term is 107, find the value of n. Also, find the 21st 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

\[S_m = 4 m^2 - m\]

\[\text{We know} \]

\[a_m = S_m - S_{m - 1} \]

\[ \therefore a_m = 4 m^2 - m - 4 \left( m - 1 \right)^2 + \left( m - 1 \right)\]

\[a_m = 8m - 5\]

\[\text{Now},\]

\[a_n = 107\]

\[\Rightarrow 8n - 5 = 107\]

\[\Rightarrow 8n = 112\]

\[\Rightarrow n = 14\]

\[a_{21} = 8\left( 21 \right) - 5 = 163\]