Question

The 11^th term and the 21^st term of an A.P. are 16 and 29 respectively, then find the 41^th term of that A.P.

Answer 1

Let a be the first term and d be the common difference.

t₁₁ = 16, t₂₁ = 29      ...(Given)

t_n = a + (n − 1)d

∴ t₁₁ = a + (11 − 1)d

∴ 16 = a + 10d 

∴ a + 10d = 16      ...(i)

Similarly, 

t_n = a + (n − 1)d

∴ t₂₁ = a + (21 − 1)d

∴ 29 = a + 20d             

∴ a + 20d = 29    ...(ii)

Subtracting equation (i) from (ii), we get,

\[\begin{array}{l}  
\phantom{\texttt{0}}\texttt{ a + 20d = 29}\\ \phantom{\texttt{}}\texttt{- a + 10d = 16}\\ \hline\phantom{\texttt{}}\texttt{ (-) (-) (-)}\\ \hline \end{array}\]

∴ 10d = 13

∴ d = `13/10`

Putting the value of d = `13/10` in equation (i),

a + 10d = 16 

∴ `a + 10(13/10) = 16`

∴ a + 13 = 16

∴ a = 16 − 13

∴ a = 3

t_n = a + (n − 1)d

∴ t₄₁ = `3 + (41 − 1)(13/10)`

∴ t₄₁ = `3 + 40 × 13/10`

∴ t₄₁ = 3 + 52

∴ t₄₁ = 55

∴ 41^st term of the A.P. is 55.