The sum of three terms of an Arithmetic Progr | vedclass

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(B) Let the three terms of the $AP$ be $(a - d)$,$a$,and $(a + d)$.Given their sum: $(a - d) + a + (a + d) = 18$.$3a = 18 \implies a = 6$.Given the su

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$11$

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(B) Let the three terms of the $AP$ be $(a - d)$,$a$,and $(a + d)$.Given their sum: $(a - d) + a + (a + d) = 18$.$3a = 18 \implies a = 6$.Given the sum of their squares: $(a - d)^2 + a^2 + (a + d)^2 = 158$.$(6 - d)^2 + 6^2 + (6 + d)^2 = 158$.$(36 - 12d + d^2) + 36 + (36 + 12d + d^2) = 158$.$108 + 2d^2 = 158$.$2d^2 = 50 \implies d^2 = 25 \implies d = \pm 5$.If $d = 5$,the terms are $(6-5), 6, (6+5)$,which are $1, 6, 11$.If $d = -5$,the terms are $(6+5), 6, (6-5)$,which are $11, 6, 1$.In both cases,the largest term is $11$.

The sum of three terms of an Arithmetic Progression $(AP)$ is $18$ and the sum of their squares is $158$. The largest term is.......

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