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Let a be the first term and d be the common difference of the A .P. We have, a 5 = 0 ⇒ a + 4d = 0 ⇒ a = – 4d Now, a 23 = a + 22d = –4d + 22d = 18d And 3 (a 11 ) = 3(a + 10d) = 3(– 4d + 10d) = 3(6d) = 18d Hence, a 23 = 3 (a 11 ) Proved.

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Arithmetic Progressions

Short Answer Type

281. Write the common difference of an A .P. whose nth term is 3n + 5.

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282. For what value of k, are the numbers x, 2x + k and 3x + 6 three consecutive terms of an A .P

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283. The 17th term of an A .P. exceeds its 10th term by 7. Find the common difference.

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284. If 9th term of an A .P. is zero. Prove that 29th term is double of its 19th term.

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285. If 5th term of an A .P. is zero. Prove that its 23rd term is three times its 11th term.

Let a be the first term and d be the common difference of the A .P.
We have,
a₅ = 0 ⇒ a + 4d = 0 ⇒ a = – 4d
Now, a₂₃= a + 22d = –4d + 22d = 18d
And 3 (a₁₁) = 3(a + 10d)
= 3(– 4d + 10d)
= 3(6d) = 18d
Hence, a₂₃ = 3 (a₁₁) Proved.

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