Question

If the 3rd term of an A.P.is 11 and the 7th term is 27, find the fifth

term of the series.​

Answer 1

Step-by-step explanation:

3rd term=11

a+2d=11.........................(1)

7th term =27

a+6d=27

Now by elimination method

a+2d=11

a+6d=27

-4d= - 16

d=4

Put the value of d in eq(1)

And get answer

Answer 2

Answer:

Here, the value of fifth term is 19.

Step-by-step explanation:

Here, As given in our question,

=T3=11, T7=27

Here, First term=a, Common difference=d,

Now, T3=a+2d=11 -(1)eq.

T7=a+6d=27-(2)eq.

=Now by applying elimination method to the equations, we get,

= a+ 2d= 11

=±a±6d=±27 (As signs of both equations are same, Considering to method sign of 2nd equation would be changed)

After solving it, we get,

=2d-6d=11-27

=(-4d) =(-16) (As minus sign is on both sides, insolving both would be cancelled)

=4d=16

=d=16/4

=d=4

By putting value of d in eq.(1), we get,

=a+2d=11

=a+2×(4)=11

=a+8=11

=a=11-8

=a=3

Now, As asked in question,

=Fifth term =T5

=T5=a+(n-1)×d (Where n=5, number of terms)

=T5=3+(5-1)×4

=T5=3+4×4

=T5=16+3

=T5=19

Hence, the value of fifth term in this A.P. is 19.

Thank you.