Assign the same string to multiple variables

In my opinion you're better of just doing the more readable:

var1="$value" var2="$value" var3="$value" var4="$value" var5="$value" var6="$value" var7="$value" var8="$value" var9="$value" var10="$value"

But if you want a very short way of accomplishing this then try:

declare var{1..10}="$value"

Edited: using brace expansions instead of printf and declare instead of eval, which could be dangerous depending on what's in $value.

Cf. EDIT1: You could still use brace expansions in the new case:

declare var{T,z,3}="$value"

It's safer than the printf approach in the comments because it can handle spaces in $value.

let is doing an arithmetic evaluation. In bash, this is equivalent to (( ... )). This is why your code only works for integers.

Using an array instead of specially named variables:

var=( "$value"  "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" )

printf 'var[5]=%s\n' "${var[5]}"

or an associative array,

declare -A var

var=( ["T"]=$value ["z"]=$value [3]=$value )

printf 'var[T]=%s\n' "${var["T"]}"

You could also do a loop:

for varname in var0 var1 varT varfoo; do
    declare -n var="$varname"
    var=$value
done

This loop loops over names of variables. In the loop body, a name reference variable is created that references the current loop variable. When the value of the name reference variable is set, the named variable is set.

value=balabala
eval var{1..10}=\$value
echo $var{1..10}