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in Arithmetic Progression by (44.4k points)
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एक समान्तर श्रेणी का 24 वाँ पद, 10 वें पद का 2 गुना है। तो सिद्ध कीजिए कि इसका 72 वाँ पद, 15 वें पद का चार गुना है।

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दिया है : 

a24 = 2 × a10

⇒ a + 23d = 2 × (a + 9d)

⇒ a + 23d = 2a + 18d

⇒ a = 5d ….(1)

अतः a72 = a + 71d

a72 = 5d + 71d [समीकरण (1) से a = 5d रखने पर]

⇒ a72 = 76d

= 4 × 19d …(2)

पुनः a15 = a + 14d

⇒ a15 = 5d + 14d

⇒ a15 = 19d …(3)

⇒ समीकरण (2) तथा (3) से

a72 = 4 × a15 यही सिद्ध करना था।

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