Question

if an = 6n + 1 , find AP , a10 and d​

Answer 1

Answer:

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Given a relation for finding the AP:

an = 6n + 1 where n denotes the number of times. Like if we want to find 1st term, then n = 1..

Now,

Putting n = 1,

➙ a1 = 6(1) + 1

➙ a1 = 7

Putting n = 2

➙ a2 = 6(2) + 1

➙ a2 = 13

Then, d = a2 - a1 because their is a common difference between any two terms of an AP. Hence,

➙ d = a2 - a1

➙ d = 13 - 7

➙ d = 6

Now, finding the AP: 7, 13, 19, 25....Keep on adding 6 to the previous term to get the next term.

To find a10, we need put n = 10,

➙ a10 = 6(10) + 1

➙ a10 = 60 + 1

➙ a10 = 61

Answer 2

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Given a relation for finding the AP:

an = 6n + 1 where n denotes the number of times. Like if we want to find 1st term, then n = 1..

Now,

Putting n = 1,

➙ a1 = 6(1) + 1

➙ a1 = 7

Putting n = 2

➙ a2 = 6(2) + 1

➙ a2 = 13

Then, d = a2 - a1 because their is a common difference between any two terms of an AP. Hence,

➙ d = a2 - a1

➙ d = 13 - 7

➙ d = 6

Now, finding the AP: 7, 13, 19, 25....Keep on adding 6 to the previous term to get the next term.

To find a10, we need put n = 10,

➙ a10 = 6(10) + 1

➙ a10 = 60 + 1

➙ a10 = 61