If a figure in the pattern we've seen has 393 squares, what figure number will it have?
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Tn=393
a=1
d=4
Tn=a+(n-1)*d
393=1+(n-1)*4
393=1+4n-4
396/4=n
99=n
n=99
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💥 If a=1 💥 But it doesn't really say a=1 There isn't a clear explanation if the figure shown is the first in the set. Omit for ambiguity.
the figure number= no. of squares on each arm+1
given, total number of squares=393; removing central square=393-1=392 number of squares on each arm=392/4=98 therefore figure number=98+1=99
4x - 3 = 393 4x = 393 +3 4x = 396 divide both sides by 4 x = 99
YEH
4n - 3 = 393
4n = 393 + 3
4n = 396
n = 396/4
n = 99
tn=393,a=1,d=4 we know that, tn=a+(n-1)d by putting the above values we get, n=99
"Un=Ua+(n-1)d"
a. Ua=1...(i);
b. d=4...(ii);
c. Un=393...(iii);
n=?
Un=Ua+(n-1)d <-(i),(ii),&(iii);
393=1+(n-1)4;
393=1+4n-4;
4n=396;
n= 4 3 9 6 =99
x= 4n-3
393= 4n-3
393+3= 4n
396 = 4n
396/4 = n
n= 99
Therefore, the position is 99th
393=4n-3 (+3)+393 = 4n 396/4 = 4n/4 (to eliminate multiple n) 99=n (so, answer is 99)
4(n)-3=393, 4(n)=393+3, n=396/4= 99
a=1 ; d=4 ; a {n} = 393. Thus, n= (a {n}-a)/d +1 implies, (393-1)/4 +1 = 98+1 =99
4n-3= 393. and 4n=393+3=396, Therefore n=396/4=99
The solution can be acquired by the formula 4n-3=393 where n is the required answer
4n-3=393 so n=99
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4n-3=393 and 4n=396,n=99