Question

In an ap sum of three consecutive terms is 27 and the product of the first and the last term is 56 find the terms​

Answer 1

Answer:

In an A.P.(Arithmetic progression),

Sum of three consecutive terma is 27.

Product of first and lats terms is 56 .

Find the terms.

$Let's \: assume \: the \: three \: terms \: as, \\ \rm (a - d), \: a, \: (a + d)(a−d),a,(a+d) respectively$

According to the question,

$\rm \implies \: (a - d) + a + (a + d) = 27 \\ \rm \implies \: a - d + a + a + d = 27 \\ \rm \implies \: a = 27$

And also,

$\rm \implies \: (a - d)(a + d) = 56 \\Substituting \: ‘a’ \\ \rm \implies \: (9 - d)(9 + d) = 56 \\ \rm \implies \: 81 - {d}^{2} = 56 \\ \rm \implies \: 81 - 56 = {d}^{2} \\ \rm \implies \: 25 = {d}^{2} \\ \rm \implies \: \sqrt{25} \\ \rm \implies \: d = 5$

Hence, the terms are :-

$\rm (a - d) = (9 - 5) = \bf4(a−d)=(9−5)=4 \\ \rm a = \bf 9a=9 \\ \rm (a + d) = (9 + 5) = \bf 14(a+d)=(9+5)=14$

Step-by-step explanation:

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Answer 2

Answer:

Hence, the terms are :-

$\begin{gathered}\rm (a - d) = (9 - 5) = \bf4(a−d)=(9−5)=4 \\ \rm a = \bf 9a=9 \\ \rm (a + d) = (9 + 5) = \bf 14(a+d)=(9+5)=14 \\ \end{gathered} </p><p>$

Step-by-step explanation:

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