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Arithmetic Progressions

Short Answer Type

291. A sum of Rs. 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these, interests form an A .P. ? If so, find the interest at the end of 30 yrs.

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292. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third and so on. There are 5 rose plant in the last row. How many rows are there in the flower bed?

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293. If m times the mth term of an A .P. is equal to n times its nth term show that (m + n)th term of the A .P. is zero.

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Long Answer Type

294. If the mth term of an A .P. is 1/n and the nth term is 1/m, find the (mn)th term.

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Short Answer Type

295. Split 69 into three parts such that they are in A .P. and the product of two smaller parts is 483.

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296. The ratio of the 7th to the 3rd term of an A .P. is 12 : 5. Find the ratio of the 13th to 4th term.

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297. If the Pth term of an A .P. is q and qth term is P, prove that its nth term is (P + q – n).

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298. Find the 10th term from the end of the A .P. 8, 10, 12................ 126

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299. The sum of the 4th and 8th terms of an A .P. is 24 and the sum of the sixth and tenth terms is 44. Find the First three terms of the A .P.

It is given that,
a₄ + a₈ = 24
⇒ a + 3d + a + 7d = 24
⇒ 2 a+ 10d = 24
⇒ a + 5d = 24 (i)
And,
a₆ + a₁₀ = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 (ii)
solving (i) and (ii) we get
a = –13 and d = 5
Hence, the first three terms are
a, a + d, a + 2d
⇒ – 13, (–13 + 5), (–13 + 2 x5)
= –13, –8,–3
⇒ 2a + 14d = 44
Problems Based on sum of ‘n’ th term of an A .P

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300.

The first and the last term of an A .P. are 4 and 81 respectively. If the common difference is 7, how many terms are there in the A .P. and what is their sum?

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