Question

The sum of three terms of an ap is 33. If the product of the first term and third term exceeds the 2th term by 29 , find the ap​

Answer 1

❍ Let's consider the first three terms of AP be a, a + d, a + 2d.

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$\bigstar\:{\underline{\boldsymbol{According\:to\:the\:question\::}}}$

• The Sum of first three terms of an AP is 33.

$:\implies\sf a + (a + d) + (a + 2d) = 33\qquad\qquad\bigg\lgroup\bf eq^n\:(1) \bigg\rgroup$

• The product of the first term and third term exceeds the 2nd term by 29.

$:\implies\sf a \times (a + 2d) = (a + d) + 29\qquad\qquad\bigg\lgroup\bf eq^n\:(2) \bigg\rgroup$

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⠀⠀⠀⠀⠀⠀⠀⠀Now, From eqⁿ (1),

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$:\implies\sf a + (a + d) + (a + 2d) = 33\\\\\\:\implies\sf 3a + 3d = 33\\\\\\:\implies\sf 3(a + d) = 33\\\\\\:\implies\sf (a + d) = \cancel{\dfrac{33}{3}}\\\\\\:\implies\sf d = 11 - a\qquad\qquad\bigg\lgroup\bf a_2 - a_1 = d \bigg\rgroup\\\\ \therefore\:{\underline{\sf{Second\:term\:of\:A.P\:is\:{\pmb{11}}.}}}$

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⠀⠀⠀⠀⠀⠀⠀⠀Now, From eqⁿ (2),

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$\qquad:\implies\sf a \times (a + 2d) = (a + d) + 29\\\\\\:\implies\sf a(a + 2(11 - a)) = 11 + 29\\\\\\:\implies\sf a(a + 22 - 2a) = 40\\\\\\:\implies\sf a(22 - a) = 40\\\\\\:\implies\sf 22a - a^2 = 40\\\\\\:\implies\sf a^2 - 22a + 40 = 0\\\\\\:\implies\sf a^2 - 20a - 2a + 40 = 0\\\\\\:\implies\sf a(a - 20) - 2(a - 20) = 0\\\\\\:\implies\sf (a - 20)(a - 2) = 0\\\\\\:\implies{\underline{\boxed{\pmb{\frak{\pink{a = 20 \:or\:2}}}}}}\:\bigstar$

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$\qquad\qquad\bf{\dag}\:\:{\underline{\textsf{By Putting a = 20 :-}}}\\\\\\ \qquad\qquad\dashrightarrow\sf d = 11 - a\\\\\\ \qquad\qquad\dashrightarrow\sf d = 11 - 20\\\\\\ \qquad\qquad\dashrightarrow\sf \purple{d = - 9}$

$\qquad\qquad\bf{\dag}\:\:{\underline{\textsf{By Putting a = 2 :-}}}\\\\\\ \qquad\qquad\dashrightarrow\sf d = 11 - 2\\\\\\ \qquad\qquad\dashrightarrow\sf d = 11 - 2\\\\\\ \qquad\qquad\dashrightarrow\sf \purple{d = 9}$

Hence, The two possible A.P's are,

• 2, 11, 20, 29, 38, 48,...
• 20, 11 , 2, -7, - 16, -25,...

Answer 2

Answer:

Solution :-

Let us assume that the first three terms are a - d,a,a + d

$\sf \: a - d + a + a + d = 33...(1)$

By cancelling d

$\sf \: a + a + a = 33$

$\sf \: 3a = 33$

$\sf \: a \: = \dfrac{33}{3}$

$\sf \: a \: = 11$

Another equation

$\sf \: (a - d)(a + d) = 29 + a$

$\sf \: {a}^{2} - {d}^{2} = 29 + a$

Putting a as 11

$\sf \: {11}^{2} - {d}^{2} = 29 + 11$

$\sf \: 121 - {d}^{2} = 29 + 11$

$\sf \: 121 - {d}^{2} = 40$

$\sf \: {d}^{2} = 121 - 40$

$\sf \: {d}^{2} = 81$

$\sf \: d = 9$

The AP is

$\therefore{ \mathfrak{AP= 20,11,2}}$