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9. Sequences and series
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Q4 of 97 Page 185

How many terms of the A.P. – 6,-, – 5, … are needed to give the sum –25?

First term a = -6


Second term = -11/2


Let d be the common difference.


d = Second term – First term


d = -11/2 – (-6) = 6 – 11/2 = 1/2


Sn = Sum of n terms of AP = -25






⇒ n2 – 25n + 100 = 0


⇒ n2 – 20n – 5n + 100 = 0


⇒ n × (n – 20) – 5 × (n – 20) = 0


⇒ (n – 20) × (n – 5) = 0


⇒ n = 20 or 5


More from this chapter

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2

Find the sum of all-natural numbers lying between 100 and 1000, which are multiples of 5.

3

In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.

5

In an A.P., if pth term is and qth term is, prove that the sum of first pq terms is 1/2 (pq +1), where p q.

6

If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Questions · 97
9. Sequences and series
1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 21 22 23 24 25 26 27 28 29 30 31 32
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