Question

The 16th term of an arithmetic progression is 53 and the 20th term is 65. find the sum of all the odd numbers in the sequence that are less than 100

please show with steps

Answer 1

Sum = 795

Explanation:

an = a + (n - 1)*d

16th Term

a_16 = a + (16 - 1) * d

a_16 = a + 15 * d

53 = a + 15 d

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a_20 = a + 19*d

65 = a + 19d

53 = a + 15d Subtract

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12 = 4d Divide by 4

3 = d

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53 = a + 15d

53 = a + 15*3

53 = a + 45

a = 53 - 45

a = 8

The first odd term in the sequence is 8 + 3 = 11

The next term is 11 + 3 = 14

The third term is 17

So the difference is 17 - 11 = 6

The last term is

L = a + (n - 1)*6

L = 11 + (n - 1)*6

I think we have to guess at what the last term is. Let n = 10

L = 11 + 9*6 It's not large enough

Try 15

L = 11 + 14*6

L = 11 + 84

L = 95 That's correct. The last term in the sequence is 95

So

The sum is

Sum = (a + L)*n/2

n = 15

Sum = (11 + 95)*15/2

Sum = (106)*15/2

Sum = 53*15/2

Sum = 795