Question

the 7th term and 13th term of an A.P.are 34 and 64 respectively.find the series term​

Answer 1

Answer:

a+6d=34

a=34-6d

Now the second one

a+12d=64

34-6d+12d=64

6d=30

d=5

Now a=34-6*5

a=4

Series 4,9,14,19,24,29,34,39,44,49,54,59,64

Answer 2

Answer:

The series is 4, 9, 14, 19........

Step-by-step explanation:

In the question,

We have,

$a_{7}=34\\and,\\a_{13} = 64$

Now, as we know that the nth term of the AP is given by,

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

where,

a = the first term

d = the common difference

and,

n = number of terms

Now,

$34=a+(7-1)d=a+6d .........(1)\\also,\\64 = a+(13-1)d = a+12d..........(2)$

On subtracting the eqn. (2) from eqn. (1) we get,

$12d - 6d = 64 - 34 = 30\\6d = 30\\d= 5$

So, on putting this in eqn. (1) we get,

$34 = a+6(5)\\a=34-30=4\\a=4$

Therefore, the series will be a, (a + d), (a + 2d), (a + 3d).....

That is 4, (4 + 5), (4 +2(5)), (4 + 3(5))..........

Hence, the series is  4, 9, 14, 19......