The sum of n and n-1 terms of an AP is 441 and 356 , respectively. If the first term of the AP is 13 and the common difference is equal to the number of terms, find the common difference of the AP.

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Arithmetic Progression Formula: The sum of the first n terms of an arithmetic progression (AP) is given by the formula: $$ S_n = \frac{n}{2} [2a + (n - 1)d] $$ where $ S_n $ is the sum of the first n terms, $ a $ is the first term, $ n $ is the number of terms, and $ d $ is the common difference.Equation for Sum of n Terms: Given that the sum of the first n terms is 441, we can write the equation: $$ 441 = \frac{n}{2} [2(13) + (n - 1)d] $$ Simplifying, we get: $$ 441 = \frac{n}{2} [26 + (n - 1)d] $$Equation for Sum of n-1 Terms: Similarly, the sum of the first n-1 terms is 356, so we can write the equation: $$ 356 = \frac{n-1}{2} [2(13) + (n - 2)d] $$ Simplifying, we get: $$ 356 = \frac{n-1}{2} [26 + (n - 2)d] $$Substitute Common Difference: We are given that the common difference $ d $ is equal to the number of terms $ n $. So we can replace $ d $ with $ n $ in our equations: For the sum of n terms: $$ 441 = \frac{n}{2} [26 + (n - 1)n] $$ For the sum of n-1 terms: $$ 356 = \frac{n-1}{2} [26 + (n - 2)n] $$Expand and Simplify Equations: Now we have two equations with one variable $ n $. We can solve these equations simultaneously to find the value of $ n $. Let's first expand and simplify both equations: For the sum of n terms: $$ 441 = \frac{n}{2} [26 + n^2 - n] $$ $$ 441 = \frac{n}{2} [n^2 + 25] $$ $$ 882 = n(n^2 + 25) $$ $$ n^3 + 25n - 882 = 0 $$ For the sum of n-1 terms: $$ 356 = \frac{n-1}{2} [26 + n^2 - 2n] $$ $$ 356 = \frac{n-1}{2} [n^2 - 2n + 26] $$ $$ 712 = (n-1)(n^2 - 2n + 26) $$ $$ n^3 - 2n^2 + 26n - n^2 + 2n - 26 = 712 $$ $$ n^3 - 3n^2 + 28n - 26 - 712 = 0 $$ $$ n^3 - 3n^2 + 28n - 738 = 0 $$Correct Expansion of Second Equation: We notice that there is a mistake in the previous step. The expansion of the second equation is incorrect. Let's correct it and expand the equation properly: $$ 712 = (n-1)(n^2 - 2n + 26) $$ $$ 712 = n^3 - 2n^2 + 26n - n^2 + 2n - 26 $$ $$ 712 = n^3 - 3n^2 + 28n - 26 $$ $$ n^3 - 3n^2 + 28n - 738 = 0 $$ This is the correct expansion of the second equation.

The sum of $\mathrm{n}$ and $\mathrm{n}-1$ terms of an AP is $441$ and $356$ , respectively. If the first term of the AP is $13$ and the common difference is equal to the number of terms, find the common difference of the AP.

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The sum of n \mathrm{n} n and n−1 \mathrm{n}-1 n−1 terms of an AP is 441441441 and 356356356 , respectively. If the first term of the AP is 131313 and the common difference is equal to the number of terms, find the common difference of the AP.

The sum of $\mathrm{n}$ and $\mathrm{n}-1$ terms of an AP is $441$ and $356$ , respectively. If the first term of the AP is $13$ and the common difference is equal to the number of terms, find the common difference of the AP.