Question

Find the 31^st term of an A.P. whose 11^th term is 38 and 6^th term is 73.

Answer 1

Given: t₁₁ = 38, t₆ = 73   

To Find: t₃₁

Solution:

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

t₁₁ = 38
∴ t₁₁ = a + (n − 1)d
∴ t₁₁ = a + (11 − 1)d
∴  38 = a + 10d    ...(i)

Similarly, in the case of  t₆ = 73   
∴ 73 = a + 5d       ...(ii)

Subtracting equation (i) from (ii)

\[\begin{array}{l} 
\phantom{\texttt{0}}\texttt{73 = a + 5d}\\ \phantom{\texttt{}}\texttt{-38 = a + 10d}\\ \hline\phantom{\texttt{}}\texttt{(-) (-) (-)}\\ \hline \end{array}\]
∴ 35 = - 5d
∴ d = `35/(-5)` 
∴ d = - 7

Substitute the value of d in equation (i), we get,
a + 10d = 38
a + 10(-7) = 38    ...(d = - 7)
a - 70 = 38
a = 38 + 70 
a = 108

∴ t₃₁ = a + (31 - 1)d
∴ t₃₁ = a + 30d
∴ t₃₁ = 108 + 30(-7)   ...(a = 108, d = - 7)
∴ t₃₁ = 108 - 210
∴ t₃₁ = - 102
∴ 31^th term = - 102.