Question

If the 10th term of an A.P is 52 and 17th term is 20 more than the 13th term, find the A.P

Answer 1

Answer:

Given a

10

=52

a

n

=a+(n−1)d

52=a+9d(1)

also, a

17

=20+a

13

a+16d=20+a+12d

4d=20

d=5(2)

From (1) and (2)

52=a+9×5

a=7

The AP is: 7,12,17,22,...

Step-by-step explanation:

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Answer 2

Answer:

Let the first term of the AP be a and the common difference be d.

According to the problem, the 10th term of the AP is 52.

Thus,

a + 9d = 52 …(i)

Again, it is said that, the 17th term of the AP is 20 more than its 13th term.

Thus,

(a + 12d) + 20 = (a + 16d) …(ii)

Solving eq.(ii), we get,

d = 5

And then substituting this value of d in eq.(i), we get,

a = 7

Thus the AP has its first term as 7 and it has a common difference 5.

Thus, he required AP is:

7 , 12 , 17 , 22 , ….

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