The sum of three terms which are in arithmetic progression is $33$ , if the product of the $1^{s t}$ and $3^{r d}$ terms exceeds the $2^{n d}$ term by $29$ , find the three terms of the A.P.

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Solution

Let $(a - r), a, (a + r)$ be the required terms of A.P.
Therefore,

Sum of terms $= 33$

$(a - r) + a + (a + r) = 33$

$3 a = 33$

$\Rightarrow a = \frac{33}{3} = 11$

Now, according to the question-

$(a - r) (a + r) = a + 29$

$a^{2} - r^{2} = a + 29$

${(11)}^{2} - r^{2} = 11 + 29$

$121 - r^{2} = 40$

$\Rightarrow r^{2} = 121 - 40$

$\Rightarrow r = \pm \sqrt{81} = \pm 9$

Case I:- $r = + 9$

$a - r = 11 - 9 = 2$

$a + r = 11 + 9 = 20$

Required A.P. $= (a - r), a, (a + r) = 2, 11, 20$

Case II:- $r = - 9$

$a - r = 11 - (- 9) = 20$

$a + r = 11 + (- 9) = 2$

Required A.P. $= (a - r), a, (a + r) = 20, 11, 2$

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The sum of three terms which are in arithmetic progression is 33, if the product of the 1st and 3rd terms exceeds the 2nd term by 29, find the three terms of the A.P.

The sum of three terms which are in arithmetic progression is $33$ , if the product of the $1^{s t}$ and $3^{r d}$ terms exceeds the $2^{n d}$ term by $29$ , find the three terms of the A.P.