Question

if the 10th term of an AP is 52 and 17th term is 20 more than its 13th term , Find the AP

Answer 1

given, a10 =52 a + 9d =52 ...........(1)
also, a17 = 20+a13 a + 16d =20+a+12d
16d-12d=20 4d=20 d = 5 putting the
value of d in eq. (1), we get, a + 9(5) =52
a +45 = 52 a = 7 hence, the required AP
is 7,12,17,.........

Answer 2

Note: We know that nth term of an AP an = a + (n - 1) * d.

(i)

Given that 10th term of an AP is 52.

⇒ a + (10 - 1) * d = 52

⇒ a + 9d = 52  

(ii)

Given that 17th term is 20 more than its 13th term.

⇒ a + (17 - 1) * d = a + (13 - 1) * d + 20

⇒ a + 16d = a + 12d + 20

⇒ 16d - 12d = 20

⇒ 4d = 20

⇒ d = 5.

Substitute d = 5 in (1), we get

⇒ a + 9d = 52

⇒ a + 45 = 52

⇒ a = 7.

Hence:

⇒ First term a = 7.

⇒ Common difference d = 5.

Therefore, the AP is 7,12,17,22....

Hope it helps!