Summation of 3n + 2 from n = 1 to n = 14 = (3(1) + 2) + (3(2) + 2) + (3(3) + 2) + . . . + (3(14) + 2) = 5 + 8 + 11 + ... + 44 ia an arithmetic progression with first term (a) = 5, common difference (d) = 3 and last term (l) = 44 and n = 14
Sn = n/2(a + l) = 14/2(5 + 44) = 7(49) = 343
Therefore, the required summation is 343.
Ur answer in attachment hope it helps you
Answer:
I think it is a maximum value.
We need to solve the speed formula for d. To do so, let's start by moving the number of the left hand side:

Square both sides to get rid of the square root:

Now plug the known value of the speed to find the distance:

So the closest answer is the last one: d=0.155km
A. 185,190 rounded to the nearest thousand is 185,000
B. 185,990 rounded = 186,000
C. 186,789 rounded = 187,000
D. 186,899 rounded = 187,000
so ur answer is B