Question

the sum of the three consecutive term in AP is 48. the product of the first and the last term is 252. find the common difference d​

Answer 1

⠀⠀⠀☯ Given that, the sum of three consecutive terms of Arithmetic Progression (AP) is 48.

And, Product of first & last term is 252.

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Let's consider that the three consecutive terms of AP be (a - d), a & a + d.

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Therefore,

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$:\implies\sf a \: \: \cancel{- d} + a + a \:\: \cancel{+ d} = 48 \\\\\\:\implies\sf 3a = 48 \\\\\\:\implies\sf a = \dfrac{\cancel{48}}{\cancel{\: 3}} \\\\\\:\implies{\boxed{\boxed{\frak{\pink{a = 16}}}}}$

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As, Product of (a - d) & (a + d) is 252.

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$:\implies\sf (a - d) (a + d) = 252 \\\\\\:\implies\sf a^2 - d^2 = 252\\\\\\:\implies\sf (16)^2 - d^2 = 252\\\\\\:\implies\sf 256 - d^2 = 252\\\\\\:\implies\sf d^2 = 256 - 252\\\\\\:\implies\sf d^2 = 4 \\\\\\:\implies{\boxed{\boxed{\frak{\pink{d = 2}}}}}$

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Now,

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• a = 16
• (a - d), 16 - 2 = 14
• (a + d), 16 + 2 = 18

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$\therefore\:{\underline{\sf{Three \ terms \ are\ {\textsf{\textbf{14, 16, 18}}}.}}}$

$\because\:{\underline{\sf{And, \ The \ Common \ difference (d) \ is \ {\textsf{\textbf{2}}}.}}}$

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$\qquad\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\: Formulas \ for \ Arithmetic \ Progression\:\bigstar}}}}$

$\begin{lgathered}\boxed{\begin{minipage}{15 em} \sf \displaystyle \bullet a_n=a + (n-1)d \\\\\\ \bullet S_n= \dfrac{n}{2} \left(a + a_n\right) \end{minipage}}\end{lgathered}$

Answer 2

$\underline{\underline{\textsf{\maltese\:\: {\red{Given :}}}}}$

☞ Sum of three consecutive terms = 48

☞ Product of first and last term = 252

$\underline{\underline{\textsf{\maltese\:\: {\red{To Find :}}}}}$

☞ Common Difference (d) = ?

$\underline{\underline{\textsf{\maltese\:\: {\red{Solution :}}}}}$

Let the three consecutive terms of the Arithmetic Progression be $\underline{\textbf{(a - d) , a and (a + d)}}$ respectively.

According to the Question,

⇒ (a - d) + a + (a + d) = 48

⇒ a - d + a + a + d = 48

⇒ 3a = 48

⇒ a = $\dfrac{48}{3}$

⇒ a = 16

$\textbf{ \therefore a = 16}$

According to the Question,

⇒ (a - d) (a + d) = 252

⇒ a² - d² = 252 ⠀‎‏‏‎‏‏⠀⠀[(a + b)(a - b) = a² - b²]

⇒ (16)² - d² = 252

⇒ 256 - d² = 252

⇒ 256 - 252 = d²

⇒ 4 = d²

⇒ √4 = d

⇒ 2 = d

$\textbf{ \therefore d = 2}$

The terms of the Arithmetic Progression are

☞ a = 16

☞ (a - d) = 16 - 2 = 14

☞ (a + d) = 16 + 2 = 18

$\underline{\underline{\textbf{ \therefore Common Difference (d) = 2}}}$

${\underline{\underline{\bf{\therefore \; The\; consecutive \; terms \; of \;the \;Arithmetic \;Progression\; are \;14\; , \;16 \; 18 \;respectively.}}}}$