How many terms of the A.P. must be taken so that their sum is 636? An arithmetic progression is shown as the sequence 9, 17, 25, and so on.

Sn = [2a + (n – 1) d]636 = [2(9) + (n – 1) 8]636 = n [9 + 4n -4]636 = 5n + 4n24n2 + 5n – 636 = 0n = 12 or n = (not possible)

Q5 of 202 Page 9

How many terms of the A.P.must be taken so that their sum is 636?

S_n = [2a + (n – 1) d]

636 = [2(9) + (n – 1) 8]

636 = n [9 + 4n -4]

636 = 5n + 4n²

4n² + 5n – 636 = 0

n = 12 or n = (not possible)

How many terms of the sequence should be taken so that their sum is zero?

How many terms are there in the A.P. whose first and fifth terms are 14 and 2 respectively and the sum of the terms is 40?

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The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?