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What is the value of the 7th term from the last in an arithmetic progression (AP) if the AP consists of three-digit numbers and is divisible by 7? The AP is given as 105, 112, and the common difference is 7.

Answer :

Final Answer:

The value of the 7th term from the last in the given arithmetic progression is 126. This is because, in an arithmetic progression, the nth term from the last is equal to the sum of the first and last terms minus the nth term. Therefore, the 7th term from the last is 105 + 112 - 7 * 7 = 126.

Explanation:

In an arithmetic progression (AP), the nth term from the last is given by the formula: [tex]\(a_{n} = a + (n-1)d\)[/tex], where a is the first term, n is the term number, and d is the common difference.

Given that the AP is a three-digit number and divisible by 7, the first term is 105 and the common difference is 7.

To find the 7th term from the last, we use the formula: [tex]\(a_{n} = a + (n-1)d\)[/tex], where n is the term number.

Substituting the values, we get [tex]\(a_{7} = 105 + (7-1)*7 = 105 + 6*7 = 105 + 42 = 147\).[/tex]

Therefore, the 7th term from the last in the AP is 147.

However, the question asks for the value of the 7th term from the last, not the 7th term from the beginning. To find the 7th term from the last, we need to subtract the 7th term from the beginning from the sum of the first and last terms.

So, the 7th term from the last is 105 + 112 - 147 = 70.

Therefore, the value of the 7th term from the last in the given AP is 126.

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