In an A.P., if the first term is 22,

Solution:

In the given problem, we need to find the number of terms of an A.P. Let us take the number of terms as n.

Here, we are given that,

$a=22$

$d=-4$

$S_{n}=64$

So, as we know the formula for the sum of n terms of an A.P. is given by,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula we get,

$S_{n}=\frac{n}{2}[2(22)+(n-1)(-4)]$

$64=\frac{n}{2}[44-4 n+4]$

$64(2)=n(48-4 n)$

$128=48 n-4 n^{2}$

Further rearranging the terms, we get a quadratic equation,

$4 n^{2}-48 n+128=0$

On taking 4 common, we get,

$n^{2}-12 n+32=0$

Further, on solving the equation for n by splitting the middle term, we get,

$n^{2}-12 n+32=0$

$n^{2}-8 n-4 n+32=0$

$n(n-8)-4(n-8)=0$

 

$(n-8)(n-4)=0$

So, we get,

$(n-8)=0$

$n=8$

Also,

$(n-4)=0$

$n=4$

Therefore, $n=4$ or 8