The sum of the first three terms of an AP is 30 and the sum of the last three terms is 36. If the first term is 9, then the number of terms is :

Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

Answer 1

Problem Statement: An arithmetic progression (AP) is defined by the following properties: the sum of its initial three terms is 30, the sum of its final three terms is 36, and its first term is $ a = 9 $. The objective is to determine the total number of terms in this AP.

Step 1: Determining the Common Difference:
The sum of the first three terms of an AP can be expressed as $ a + (a+d) + (a+2d) $. Given that $ a = 9 $ and this sum equals 30, we have:
\[9 + (9 + d) + (9 + 2d) = 30\]Simplifying the equation:
\[27 + 3d = 30\]\[3d = 3\]\[d = 1\]Therefore, the common difference $ d $ is 1.

Step 2: Calculating the Number of Terms:
Let $ n $ be the total number of terms. The last three terms of the AP are $ a + (n-3)d $, $ a + (n-2)d $, and $ a + (n-1)d $. Their sum is given as 36. Substituting $ a = 9 $ and $ d = 1 $:
\[(9 + (n-3)(1)) + (9 + (n-2)(1)) + (9 + (n-1)(1)) = 36\]Simplifying this expression:
\[(9 + n - 3) + (9 + n - 2) + (9 + n - 1) = 36\]\[(n+6) + (n+7) + (n+8) = 36\]\[3n + 21 = 36\]\[3n = 15\]\[n = 5\]Thus, the AP contains 5 terms.

Conclusion:
The arithmetic progression has 5 terms.

The sum of the first three terms of an AP is 30 and the sum of the last three terms is 36. If the first term is 9, then the number of terms is :

The Correct Option is B

Solution and Explanation

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