Question

The 11th term of an ap is 80 and the 16th term is 110. find the 31st term

Answer 1

a+10d=80 and a+15d=110.subtracting both equation so we get d=6. and a=20.now its 31term is t31=20+30*6=200.

Answer 2

Given,

11th term of A.P. = 80

16th term of A.P. = 110

To Find,

31st term of A.P. =?

Solution,

We know that the nth term of AP is  $a_n = a+(n-1)d.$

Therefore,

$a_{11} = a+(11-1)d. = a +10d = 80$ ⇒ Equation 1

$a_{16} = a+(16-1)d. = a +15d = 110$⇒ Equation 2

Subtracting eq1 from eq 2, we get

$a+ 15d - a -10d = 110 -80$

$5d = 30$

$d = 30/5 =6$

Therefore, the common difference(d) of AP is 6

Putting the value of d in eq1,

$a + 10*6 = 80$

$a = 80- 60$

a = 20

Therefore, the first number in AP  is 20

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

Putting values of a and d in this equation

$a_{31} = 20 + 30*6$

$a_{31} = 20 +180$

$a_{31} = 200$

Hence, the 31st term of the given AP is 200.