Question

If rhe sum of three consecutive terms of ap is 48 and product of first and last term is 252

Answer 1

Let the numbers be ,

x-d , x , x +d

sum of them = 48

3x = 48

x = 16

now,

product of first and last = 252 (given)

(16-d)(16+d) = 252

256-d² = 252

d = 2

so, numbers are 14, 16, 18 (Ans)

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

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

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