Question

Find the number of terms of the AP $18,15,12,$___ so that their sum is $45$. Explain the double answer.

• A. $n=3,10$
  
• B. $n=2,10$
  
• C. $n=N$
  
• D. $n=10$
  

Answer 1

The correct option is A $n=3,10$

$18,15,12,..........$

$S_n=45=\frac{n}{2}[2a+(n-1\left)d\right]$

$=>\frac{n}{2}[2\times 18+(n-1\left)\right(-3\left)\right]=45$

$=>\frac{n}{2}[36-3n+3]=45$

$=>n(39-3n)-90=0$

$=>39n-3n^2-90=0$

$=>3n^2-39n+90=0$

$=>n^2-13n+30=0$

$=>n^2-10n-3n+30=0$

$=>n(n-10)-3(n-10)=0$

$=>(n-10)(n-3)=0$

$=>n=10$ or $n=3$

We can say numbers in the series are descending we will get negative values.

When, $n=3$ series is

$18,15,12$

$\because S_3=45$

When $n=10$ series is

$18,15,12,9,6,3,0,-3,-6,-9$

$\because S_{10}=45$