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How many terms of the A.P. $45, 39, 33, …,$ must be taken so that their sum is 180?Explain the double answer.
Given:
Given A.P. is $45, 39, 33, …,$
To do:
We have to find the number of terms that must be taken so that their sum is 180.
Solution:
Let the number of terms be \( n \).
First term \( (a)=45 \)
Common difference \( (d)=39-45=-6 \)
We know that,
\( S_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( \Rightarrow 180=\frac{n}{2}[2 \times 45+(n-1) \times(-6)] \)
\( \Rightarrow 180=\frac{n}{2}[90-6n+6] \)
\( \Rightarrow 360=n(-6 n+96) \)
\( \Rightarrow 6\times60=6(- n^{2}+16 n) \)
\( \Rightarrow n^{2}-16 n+60=0 \)
\( \Rightarrow n^{2}-10 n-6 n+60=0 \)
\( \Rightarrow n(n-10)-6(n-10)=0 \)
\( \Rightarrow(n-10)(n-6)=0 \)
This implies,
\( n-10=0 \) or \( n-6=0 \)
\( n=10 \) or \( n=6 \)
The number of terms to be taken is 6 or 10. 
This is because as the common difference is $-6$ the A.P. turns from positive to negative after few terms which makes the sum of the terms from 7th term to 10th term $0$.
$a_7=a+6d=45+6(-6)=45-36=9, a_8=a_7+d=9-6=3, a_9=a_8+d=3-6=-3, a_10=a_9+d=-3-6=-9$
Here,
$a_7+a_8+a_9+a_10=9+3-3-9=0$
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