Question

For a sequence of natural numbers, find S50 – S40

Answer 1

Step-by-step explanation:

ANSWER

We know the sum of n terms of an A.P with first term a and the common difference d is:

S

n

=

2

n

[2a+(n−1)d]

Since the sequence is of natural numbers.

We find S

50

with the first term is a=1, common difference is d=2−1=1 and n=50, therefore, the sum is:

S

50

=

2

50

[(2×1)+(50−1)(1)]=25(2+49)=25×51=1275

Now, we find S

40

with the first term is a=1, common difference is d=2−1=1 and n=40, therefore, the sum is:

S

40

=

2

40

[(2×1)+(40−1)(1)]=20(2+39)=20×41=820

Thus, S

50

−S

40

=1275−820=455

Hence, S

50

−S

40

=455.