Question

find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.​

Answer 1

Sølûtíoñ=>

11th term=a+10d

16th term=a+15d

now,

a/c to question,

a+10d=38-----------------(1)

a+15d=73-----------------(2)

on subtracting eq(2) from eq(1)

we get, d=7

:putting the value of d in eq (1)

a+10d=38

a=38-10*7

a=38-70

a=-32

now, 31st term=a+30d

put the value of a and d

so,a+30d

=>-32+30*7

=>-11

therefore, the 31st term is -11

✌️Hópé ît helps ú✌️

Answer 2

SOLUTION:-

Given:

⚫11th term is 38.

⚫16th term is 73.

Using Formula of Arithmetic Progression;

${}^{a} n = a + (n - 1)d$

11th term of an A.P.;

${}^{a} 11 = a + (11 - 1)d \\ \\ = > 38 = a + 10d \\ \\ = > a = 38 - 10d.............(1)$

&

16th term of an A.P.is 73.;

${}^{a} 16 = a + (16 - 1)d \\ \\ = > 73 = a + 15d \\ \\ = > a = 73 - 15d...............(2)$

From equation (1) & (2), we get;

=) 38- 10d =73 - 15d

=) -10d + 15d = 73 -38

=) 5d = 35

=) d= 35/5

=) d= 7

Putting the value of d in equation (1), we get;

=) a= 38 -10d

=) a= 38 - 10(7)

=) a= 38 - 70

=) a= -32

Now,

31th term of an A.P.

${}^{a} 31 = a + (31 - 1)d \\ \\ = > - 32 + 30(7) \\ \\ = > - 32 + 210 \\ \\ = > 178$

Thus,

The 31th term of A.P. is 178.

Hope it helps ☺️