Question

how many terms of the ap 3,6,9,12...should be taken so that the sum is 165
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Answer 1

Given-

AP = 3,6,9,12....

a(first term) = 3

d(common difference) = 3

To find-

number of terms which sum equal to 165

Solution-

Sn = n/2(2a +(n-1)d)

165 = n/2(6 + 3n -3)

330 = 3n^2 + 3n

cancelling 3 we get

n^2 + n -110 = 0

n^2 + 11n - 10n - 110 = 0

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

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

n = -11 , n=10

n cannot be negative so n=10terms

So n=10terms is required answer.

Answer 2

Concept

The sum of an AP is

Sum = n/2( 2a + (n-1)d)

where a = first term

d = common difference

n = number of terms

Given

an AP 3,6,9,12...

Find

we need to find the number of terms of the given AP for which its sum is 165.

Solution

We have

3,6,9,12....

Here, a = 3,

d = 6-3 = 3

Thus, the sum will be given by

165 = n/2(2a + (n-1)d)

165 = n/2(6 + (n-1)*3)

330 = n(6 + 3n- 3)

330 = 6n + 3n^2 -3n

330 = 3n - 3n^2

3n^2 - 3n + 330 = 0

n^2 - n +110 = 0

n^2 + 11n - 10n - 110 = 0

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

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

n = -11 , n=10

as n cannot be a negative number, n = 10

Thus, the number of terms for which the given AP's sum is 165 is 10.

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