Question

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

Answer 1

Answer:

a31=178

Step-by-step explanation:

$a11 = 38 \\ a16 = 73 \\ a + 10d = 38 \\ a + 15d = 73 \\ - 5d = - 35 \\ d = 7 \\ a + 10(7) = 38 \\ a = - 32 \\ an = a + (n - 1)d \\ a31 = - 32 + 30(7) \\ a31 = - 32 + 210 \\ a31 = 178$

Answer 2

Answer:

 

178

Step-by-step explanation:

$a_{31} = ?$

$a_{11} = 38$

$a_{16} = 73$

Now,

a + (11 - 1) d = 38

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

and,

a + (16 - 1 ) d = 73

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

By (2) - (1) , we get :-

5 d = 35

d = 7 -------------------------------- (3)

Now, putting value of d from (3) in (1) , we get,

a + 10(7) = 38

a + 70 = 38

a = 38 - 70

a = - 32

We can also verify the values of 'a' and 'd' by putting them in any of the above equation. For instance :-

$a_{11} = 38$

So,

-32 + 10(7)

-32 + 70

38 = R.H.S. [Hence the values are correct]

Now,

$a_{31} =$ a + (31 - 1)d

                = a + 30d

                = -32 + 30(7)

                = -32 + 210

          = 178

Thanks !

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