Question

In an $A.P.$ the 6th term is half the $4^{th}$ term and $3^{rd}$ term is $15$. How many terms are needed to give a sum that is equal to $66$?

Answer 1

We have,

In an A.P.

Sixth term$=\frac{fourthterm}{2}$

Let the first term $=a$

Common difference$=d$

Then,

$T_6=\frac{T_4}{2}$

$a+5d=\frac{a+3d}{2}[\therefore a_n=a+(n-1)d]$

$\Rightarrow 2a+10d=a+3d$

$\Rightarrow 2a-a=3d-10d$

$\Rightarrow a=-7d$

Now, According to given question,

Third term = 15

$a+2d=15$

$\Rightarrow -7d+2d=15$

$\Rightarrow -5d=15$

$\Rightarrow d=-3$

So,

$a=21$

Now,

Sum of n terms$=66$

$\frac{n}{2}[2a+(n-1)d]=66(\therefore S_n=\frac{n}{2}(2a+(n-1)d))$

$\Rightarrow \frac{n}{2}[2\times 21+(n-1)(-3)]=66$

$\Rightarrow \frac{n}{2}[42-3n+3]=66$

$\Rightarrow \frac{n}{2}[45-3n]=66$

$\Rightarrow \frac{3n}{2}[15-n]=66$

$\Rightarrow \frac{n}{2}(15-n)=22$

$\Rightarrow n(15-n)=44$

$\Rightarrow 15n-n^2=44$

$\Rightarrow n^2-15n+44=0$

$\Rightarrow n^2-(11+4)n+44=0$

$\Rightarrow n^2-11n-4n+44=0$

$\Rightarrow n(n-11)-4(n-11)=0$

$\Rightarrow (n-11)(n-4)=0$

If, $n-4=0,n=4$

If, $n-11=0,n=11$

So, after 4th term the sum of all terms will be zero

Hence, this is the answer.