Question

How do you find the sum of the arithmetic sequence 100 + 95 + 90 + ... 5?

Answer 1

Here a1=100

d=95-100

d=-5

last term (l)=5

l=a+(n-1)d

5=100+(n-1)×-5

5-100=(n-1)×-5

-95/-5=n-1

19+1=n

n=20

Sn=n/2(a+l)

Sn=20/2(100+5)

Sn=10×105

Sn=1050

SUM OF ARITHMETIC SEQUENCE IS EQUAL TO 1050

Answer 2

GIVEN :-

a = 100

d = -5

l = 5

TO FIND :- Sum of the AP

SOLUTION :-

${a}_{n} = a + (n - 1)d \\ \\ \implies 5 = 100 + (n - 1) - 5 \\ \\ \implies - 95 = (n - 1) - 5 \\ \\ \implies 19 = n - 1 \\ \\ \implies \boxed{n = 20}$

Now, we shall find the sum of the AP,

${s}_{n} = \frac{n}{2} (a + l) \\ \\ = \frac{20}{2} (100 + 5) \\ \\ = 10 \times 105 = 1050(ans)$

The sum of the sequence is 1050.