Question

the fifth term of an arithmetic progression is 40 and the seventh term is 28 more than the 3rd term. Find the first and 10th term.​

Answer 1

Explanation:

t5 = a + (5-1)d

t5 = a + 4d

40 = a + 4d

t3 = a + 2*d

t7 = a + 2d + 28 = a + 6d

Start with the seventh term which has 2 equal answers

a + 2d + 28 = a + 6d Subtract a from both sides

2d + 28 = 6d Subtract 2d from both sides

28 = 4d Divide by 4

28/4 = d

d = 7

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t5 = a + 4d = 40

a + 4*7 = 40

a = 40 - 28

a = 12

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First term is 12

10th term is

t10 = 12 + (10 - 1)*7

t10 = 12 + 9*7

t10 = 75

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