Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour

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


Answer :

Given \(a_{11} = 38 \) and \(a_{16} = 73\)

Using formula \(a_n = a + ( n - 1)d \), to find nth term of arithmetic progression,

\(\Rightarrow \) 38 = a + (11 - 1)(d), and
73 = a + (16 - 1)(d)
\(\Rightarrow \) 38 = a + 10d, and
73 = a + 15d

These are equations consisting of two variables.

We have,
\(\Rightarrow \) 38 = a + 10d
\(\Rightarrow \) a = 38 - 10d
Let us put value of a in equation
\(\Rightarrow\) (73 = a + 15d),
\(\Rightarrow \) 73 = 38 - 10d + 15d
\(\Rightarrow \) 35 = 5d

Therefore, Common difference = d = 7

Putting value of d in equation
\(\Rightarrow \) 38 = a + 10d,
\(\Rightarrow \) 38 = a + 70
\(\Rightarrow \) a = -32

Therefore, common difference = d = 7 and
First term = a = –32

Using formula \(a_n = a + ( n - 1)d \), to find nth term of arithmetic progression,

\(\Rightarrow a_{31}\)= -32 + (31 - 1) (7)
\(\Rightarrow \) -32 + 210 = 178

Therefore, 31st term of AP is 178.

NCERT solutions of related questions for Arithmetic Progressions

NCERT solutions of related chapters class 10 maths

NCERT solutions of related chapters class 10 science