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Home » The sum of three terms of an A.P. is and the product of the first and the third terms exceeds the second term by , find three terms.
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
Arithmetic Progressions | CBSE | Class 10 | Exercise 9.5 | Maths | RD Sharma
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

Let’s consider the three terms of the A.P. to be a-d, a, a+d

Therefore, the sum of three terms =21

⇒a-d+a+a+d=21

⇒3a=21

⇒a=7 It is already given the product of the first and {{3}^{rd}}term={{2}^{nd}}term+6

Therefore,

⇒(a-d)(a+d)=a+6

{{a}^{2}}-{{d}^{2}}=a+6

\Rightarrow {{\left( 7 \right)}^{2}}-{{d}^{2}}=7+6

\Rightarrow {{d}^{2}}=49-13=36

\Rightarrow {{d}^{2}}={{\left( 6 \right)}^{2}}

\Rightarrow d=6

Hence, the terms are 7-6, 7, 7+6 ⇒ 1, 7, 13

 

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