Question 7
Find the $31^{st}$ term of an A.P. whose $11^{th}$ term is 38 and the $16^{th}$ term is 73
Solution:
Given that,
$11^{th}$ term, $a_{11}$ = 38
and $16^{th}$ term, $a_{16}$ = 73
We know that,
| $a_n$ = a+(n−1)d |
$a_{11}$ = a+(11−1)d
38 = a+10d ………………………………. (i)
In the same way,
$a_{16}$ = a +(16−1)d
73 = a+15d ………………………………………… (ii)
On subtracting equation (i) from (ii), we get
35 = 5d
d = 7
From equation (i), we can write,
38 = a+10×(7)
38 − 70 = a
a = −32
$a_{31}$ = a +(31−1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, $31^{st}$ term is 178