Question

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

Answer 1

Given,

T11 = 38

T16 = 73

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

a + 15d= 73 --------(2)
________________subtract eq(1) from eq (2), we get ,



5d = 35


d = 7


put value of "d" in (1) , we get

a + 10 (7) = 38



a = 38 - 70 = - 32


a = - 32


we know T31 = a + 30 d

so now ,



T31 = -32 + 30 ( 7 )



T31 = - 32 + 210


T31 = 178

hence, T31 = 178


【 hope it helps 】

Answer 2

Solution:

Given:

$\sf{\implies a_{11}=38}$

$\sf{\implies a_{16}=73}$

To Find:

=> 31st term

Formula used:

$\sf{\implies a_{n}=a+(n-1)d}$

So, we know that

$\sf{\implies a_{11}=a+10d}$

$\sf{\implies a+10d=38\;\;\;..........(1)}$

$\sf{\implies a_{16}=a+15d}$

$\sf{\implies a+15d=73\;\;\;..........(2)}$

By using substitution method we will find value of a and d.

=> a + 10d = 38      .......(1)

=> a + 15 d = 73     .......(2)

=> a + 10d = 38

=> a = 38 - 10d

Put the value of a in Equation (2), we get

=> a + 15d = 73

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

=> 38 + 5d = 73

=> 5d = 73 - 38

=> 5d = 35

=> d = 35/5

=> d = 7

Put the value of d in Equation (1), we get

=> a + 10d = 38

=> a + 70 = 38

=> a = 38 - 70

=> a = -32

Now, we will find 31st term,

$\sf{\implies a_{31}=a+(n-1)d}$

=> -32 + (31 - 1)7

=> -32 + (30)7

=> -32 + 210

=> 178

So, 31st term of an AP is 178.