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S n = 5n 2 + 3n ⇒ S n – 1 = 5(n – 1) 2 + 3(n – 1) ∴ S n – S n – 1 = (5n 2 + 3n) – [5(n – 1) 2 + 3(n – 1)} ⇒ a n = (5n 2 + 3n) – {5 (n 2 – 2n + 1) + 3n – 3} = (5n 2 + 3n) – {5n 2 – 10n + 5 + 3n –3} = 5n 2 + 3n – 5n 2 + 7n – 2 = 10n – 2 Hence, required n th term is 10n – 2

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Arithmetic Progressions

Short Answer Type

301. If the sum of first 7 terms of an A .P. is 49 and that of first 17 terms is 289, find the sum of first n terms.

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302. If S_n, the sum of first n terms of an A .P. is given by S_n = 3n² – 4n, then find its n^thterm.

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303.

If S_n, the sum of first n terms of an A .P. is given by S_n = 5n² + 3n, then find its n^thterm.
S_n = 5n² + 3n

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304. Find the sum of the :
First 25 terms of an A .P. whose nth term is given by a_n= 7 – 3n. z

S_n = 5n² + 3n
⇒ S_{n – 1} = 5(n – 1)² + 3(n – 1)
∴ S_n – S_{n – 1} = (5n² + 3n) – [5(n – 1)² + 3(n – 1)}
⇒ a_{n_{= (5n²+ 3n) – {5 (n² – 2n + 1) + 3n – 3}}}= (5n² + 3n) – {5n² – 10n + 5 + 3n –3}
= 5n² + 3n – 5n² + 7n – 2
= 10n – 2
Hence, required n^th term is 10n – 2

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305.

Find the sum of the :
(i) First 25 terms of an A .P. whose nth term is given by a_n= 7 – 3n

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306. The sum of n terms of 2 A .P.’s sire in the ratio 7n + 1 : 4n + 27, show that the ratio of their 11th term is 4 : 3.

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307. If the sum of m terms of an A .P. is the same as the sum of its nth terms. Show that the sum of its (m + n)th terms is zero.

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308. If S₁, S₂,S₃, be the sum of n, 2n and 3n terms S₃ = 3(S₁ – S₁).

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Long Answer Type

309.

The sum of the first p, q and r terms of an A .P. are a, b, c respectively.
Prove that $straight a over straight p left parenthesis straight q minus straight r right parenthesis plus straight b over straight q left parenthesis straight r minus straight p right parenthesis plus straight c over straight r left parenthesis straight p minus straight q right parenthesis equals 0$