Wednesday 26 December 2012

CBSE I NCERT Arithmetic Progression-X Solved Problems


1. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top .If the top and the bottom rungs are two and a half meter apart, what is the length of the wood required for the rungs?

2. The sum of first n terms of an AP is given by Sn = 3n2 + 5n find the nth term of the AP.

3. How many terms of the ap -6,-11/2 , -5,...... are needed to give the sum -25

4. Find a, b such that 27, a, b - 6 are in A.P.

5. Find the sum of all the odd numbers between 50 and 150 divisible by 7

6. The sum of third and seventh term of an AP is 6 and their product is 8.

7. If Sn = n2p and Sm = m2p, (m not equal ton), is an A.P. prove that Sp = p3.

8.Show that the sum of (m+n)th term and (m-n)thterm of an A.Pis equal to twice the mth term.

9. If the ratio of the sums of n terms of 2 APs is n+1:3n+1, then find the ratio of the 7th terms of the AP.

10. Determine the sum of the first 30 terms of the sequence whose nth term is given by tn=2n+9/3

11. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

12.The sum of the first and the last terms of an AP is 60,the sum of n terms of the AP is720.what is n?

13. If pth , qth and rth term of an AP are a,b,c respectively , then show that (a-b)r +(b-c)p + (c-a)q = 0

14. if (b+c)/a, (c+a)/b, (a+b)/c are in A.P. show that bc,ca,ab re in A.P.

15. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms

Class X ARITHMETIC PROGRESSIONS (8) Periods

Motivation for studying AP. Derivation of standard results of finding the nth term and sum of first n terms and their application in solving daily life problems.

2 comments:

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