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Narayan singh Best Answer
Let a1, a2.....an an A.P., whose common difference is d. Now, pth term from the beginning, =) ap= a1+(p-1)d pth term from the last =)(n-p+1)the term from the beginning. =) an-p+1 =) a1+(n-p+1-1)d =) a1+ (n-p)d Therefore, (pth term from the beginning)+(pth term from the end) =)[a1+(p-1)d]+[a1+(n-p)d] =) 2a1+ (n-1)d =) a1+ [a1+(n-1)d] =) a1+ an =) sum of first and last term.