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

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