Question

find the 31st term of an AP whose 10th term is 38 and 16 th term is 74

Answer 1

$\underline{\bold{Answer:-}}$


Let first term be a
and common difference be d


A / q

a10 = 38

=> a + 9d = 38 ---------- ( 1 )

And

a16 = 74

=> a + 15d = 74 ----------- ( 2 )

On solving ( 2 ) from ( 1 )

=> a + 15d = 74

=> a + 9d = 38
( - )--( - )----( - )
____________

=> 6d = 36

=> d = 6

From ( 1 )

a + 9d = 38

=> a + 9 × 6 = 38

=> a + 54 = 38

=> a = 38 - 54

=> a = - 16

Here a = - 16 , d = 6


We know that nth terms of an AP is given by :-


an = a + ( n - 1 ) × d

a31 = - 16 + ( 31 - 1 ) × 6

a31 = - 16 + 30 × 6

a31 = - 16 + 180

a31 = 164.

Hence, the 31st term of an AP is 164.









$\textbf{Hope it helps!}$

Answer 2

a10=38=> a+9d=38....(1)
a16=74=>a+15d=74....(2)
Solving (1) &(2) for a and d
subtracting (1) from (2)
6d=36
d=36/6
d=6

on putting the value of d in (1)
a+9×6=38
a+54=38
a=38-54= -16

Now a31=a+30d
= -16 + 30×6
= -16+180
=164