Question

If the 4th and 11th term of a AP are 21 and 35.Find the 16th term of the progression?

Answer 1

4th term = 21
a + 3d = 21
a = 21 - 3d -----: ( 1 )



11th term
a + 10d = 35
a = 35 - 10d

From ( i ),

21 - 3d = 35 - 10d
10d - 3d = 35 - 21
7d = 14
d = 2


Putting value in ( i ),
a = 21 - 3d
a = 21 - 3( 2 )
a = 21 - 6
a = 15






Then, 16th term = a + 15d
=> 15 + 15( 2 )
=> 15 + 30
=> 45
$.$

Answer 2

Given,

T4=21

a+3d=21

a=21-3d................eq(i)

And,

T11=35

a+10d=35

(21-3d)+10d=35..........from eq(i)

21+7d=35

7d=35-21

7d=14

d=2

putting the value of d in eq(i)

a=21-3×2

a=15

Hence , T16=a+15d

=15+15×2

=15+30

=45

hope this helps u....

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