Unknown. C is the hypotenuse, but this problem is solvable.
A. 6082
B.120687
C.1188228
D.19042587
Step-by-step explanation:
∑k=150(3k+2)
First find a1 and a50 :
a1=3(1)+2=5a20=3(50)+2=152
Then find the sum:
Sk=k(a1 + ak)2S50=50(5 + 152)2=3925
Here is your answer
Money saved on each successive day is-
$3, &5, $7....
Clearly it forms an AP,
where
a1= 3
common difference, d= 2
n=20
So,
using formula
Tn= a1+(n-1)d
T20= 3+(20-1)2
= 3+ 19×2
= 3+ 38
=41
So, money saved on August 20= $41
Sum of money saved upto August 20 =
n/2 (a1+T20)
= 20/2 (3+41)
= 10× 44
= $440
HOPE IT IS USEFUL
D is the answer
6*10*3= 180 or
6*10 = 60*3 = 180
