How many terms of the AP 3 6 9 must be added to get the sum 165?

The number of terms of the arithmetic progression (AP) 3, 6, 9... that must be added to get a sum of 165 is 10.

Step-by-Step Explanation

Here's how we arrive at that answer, drawing upon the understanding of arithmetic progressions and the given information:

The AP is given by: 3, 6, 9, ...

• First term (a): 3
• Common difference (d): 6 - 3 = 3

We want to find the number of terms (n) such that the sum (S_n) is 165.

The formula for the sum of an AP is:

S_n = (n/2) * [2a + (n-1)d]

Substituting the known values:

165 = (n/2) [2(3) + (n-1)3]
165 = (n/2) [6 + 3n - 3]
165 = (n/2) [3n + 3]
165 = (n/2) 3(n + 1)
165 = (3n/2) * (n + 1)
330 = 3n(n + 1)
110 = n(n + 1)
n² + n - 110 = 0

Now, we need to solve this quadratic equation for n.

Factoring the quadratic equation:

(n - 10)(n + 11) = 0

Therefore, the possible values for n are 10 and -11. Since the number of terms cannot be negative, we discard -11.

Therefore, n = 10.

Conclusion:

Therefore, the number of terms in the AP that must be added to obtain a sum of 165 is 10, confirming the information from the provided reference.