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Determine k so that k2 + 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP. - Mathematics

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प्रश्न

Determine k so that k2 + 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP.

बेरीज

उत्तर

Since, k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are consecutive terms of an AP.

∴ 2k2 + 3k + 6 – (k2 + 4k + 8)

= 3k2 + 4k + 4 – (2k2 + 3k + 6) ...[Common difference]

⇒ 2k2 + 3k + 6 – k2 – 4k – 8 = 3k2 + 4k + 4 – 2k2 – 3k – 6

⇒ k2 – k – 2 = k2 + k – 2

⇒ – k = k

⇒ 2k = 0

⇒ k = 0

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पाठ 5: Arithematic Progressions - Exercise 5.3 [पृष्ठ ५३]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
पाठ 5 Arithematic Progressions
Exercise 5.3 | Q 11 | पृष्ठ ५३
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