Question

If the 11th term and 16th term of an AP is 38 and 73 repectively, then find its 31st term.

• A. 31
• B. 0
• C. 178
• D. 100

Answer 1

The correct option is C 178

Given, the 11th term and 16th term of an AP is 38 and 73 repectively.
$\because$ nth term of an AP = a + (n-1)d
$\therefore$ 11th term = a + (11-1)d
= a + 10d
$\Rightarrow$ a + 10d = 38............(i)
16th term = a + (16-1)d
= a + 15d
$\Rightarrow$ a + 15d = 73...........(ii)
Subtracting (i) from (ii), we get,
a + 15d - (a + 10d) = 73 - 38
$\Rightarrow$ 5d = 35
$\Rightarrow$ d = 7
Substituting 'd' in (i), we get,
a + 10(7) = 38
$\Rightarrow$ a = 38 - 70
= - 32
$\therefore$ 31st term of the AP = a + (31-1)d
= - 32 + (30)7
= -32 + 210
= 178