The first term of an A.P. of consecutive inte | vedclass

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(D) Given that the $A.P.$ consists of consecutive integers,the common difference $d = 1$.The first term $a = p^2 + 1$.The number of terms $n = 2p + 1$

Vedclass · Accepted Answer

$p^3 + (p + 1)^3$

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(D) Given that the $A.P.$ consists of consecutive integers,the common difference $d = 1$.The first term $a = p^2 + 1$.The number of terms $n = 2p + 1$.The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values:$S_{2p+1} = \frac{2p + 1}{2} [2(p^2 + 1) + (2p + 1 - 1)(1)]$$S_{2p+1} = \frac{2p + 1}{2} [2p^2 + 2 + 2p]$$S_{2p+1} = (2p + 1)(p^2 + p + 1)$We know that $(p + 1)^3 = p^3 + 3p^2 + 3p + 1$.Expanding $p^3 + (p + 1)^3 = p^3 + p^3 + 3p^2 + 3p + 1 = 2p^3 + 3p^2 + 3p + 1$.Alternatively,multiplying $(2p + 1)(p^2 + p + 1) = 2p^3 + 2p^2 + 2p + p^2 + p + 1 = 2p^3 + 3p^2 + 3p + 1$.Thus,the sum is $p^3 + (p + 1)^3$.

The first term of an $A.P.$ of consecutive integers is $p^2 + 1$. The sum of $(2p + 1)$ terms of this series can be expressed as:

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