Question

Find the \(31^{st}\) term of an AP whose \(11^{th}\) term is 38 and the \(16^{th}\) term is 73.

Answer 1

Given that, \(a_{11} = 38\) and \( a_{16} = 73\)
We know that, 
\(a_n = a + (n − 1) d\)
\(a_{11} = a + (11 − 1) d \)
\(38 = a + 10d    \)    ……..(1) 
Similarly,
\(a_{16} = a + (16 − 1) d\)
\(73 = a + 15d  \)    …….(2)
On subtracting (1) from (2), we obtain
\(35 = 5d\)
\(d = 7\)
From equation (1),
\(38 = a + 10 × (7) \)
\(38 − 70 = a \)
\(a = −32\)
\(a_{31} = a + (31 − 1) d\)
\(a_{31}= − 32 + 30 (7) \)
\(a_{31}= − 32 + 210\)
\(a_{31}= 178\)

Hence, \(31^{st}\) term is \(178\).