Question

Find the ap whose tenth term is 52 and whose 17 term exceeds its 13 term by 20

Answer 1

refer to the attachment

Answer 2

Answer:

$\large\boxed{\sf{7,12,17,22,........}}$

Step-by-step explanation:

Let the first term is "a"

And the common difference is "d"

According to Question:

$a +9d = 52$

And

$a + 12d + 20 = a + 16d \\ \\ = > 16d - 12d = 20 \\ \\ = > 4d = 20 \\ \\ = > d = \frac{20}{4} \\ \\ = > d = 5$

Therefore, we will get

$a + 9 \times 5 = 52 \\ \\ = > a + 45 = 52 \\ \\ = > a = 52 - 45 \\ \\ = > a = 7$

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