Question

If A = 1 + 3 + 5 + ...... +99; B = 2 +4 +6 + 8 + ........ +100 then the value of A-B is​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Concept:

The formula for finding the nth term of an Arithmetic progression (AP) is

$a_{n} = a + (n - 1)d$

where an is the nth term

a is the 1st term

and d is the common difference

The formula for the sum of an AP is given by

$s_{n} = \frac{n}{2} (2a + (n - 1)d)$

Given:

2 series A and B such that

A=1+3+5+.....+99

B=2+4+6+8+......+100

Find:

we need to find the value of A-B

Explain:

We have,

A=1+3+5+...+99

The above series is an AP with a= 1 and d=2

$a_{n} = a + (n - 1)d$

99 = 1+ (n-1)2

99-1 = (n-1)2

98 = (n-1)2

49 = n-1

n = 50

Now, the sum of the AP will be

$s_{n} = \frac{n}{2} (2a + (n - 1)d)$

A = 50/2 [2+ (50-1) 2]

A = 25 [2+ 98]

A = 25 x 100

A = 2500

Similarly,

B=2+4+6+8+......+100

The above series is an AP with a= 2 and d= 2

$a_{n} = a + (n - 1)d$

100 = 2+ (n-1) 2

100 -2 = (n-1)2

98 = (n-1)2

49 = (n-1)

n = 50

Now, the sum of the AP will be

$s_{n} = \frac{n}{2} (2a + (n - 1)d)$

B = 50/2 [(22) + (50-1) 2 ]

B = 25 (4+98)

B = 25 x 102

B = 2550

Therefore, A-B = 2500 -2550 = -50

Thus the value of A-B is -50

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