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FIND THE SUM OF 20 TERMS OF THE ARITHMETIC SERIES IN WHICH 3rd TERM IS 7 AND 7th TERM IS 2 MORE THAN THREE TIMES ITS 3rd TERM

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Answer:

740

Explanation:

The n th term of an arithmetic series is


a_(n) = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 7 and a₇ = (3 × 7) + 2 = 21 + 2 = 23 , then

a₁ + 2d = 7 → (1)

a₁ + 6d = 23 → (2)

Subtract (1) from (2) term by term

4d = 16 ( divide both sides by 4 )

d = 4

Substitute d = 4 into (1)

a₁ + 2(4) = 7

a₁ + 8 = 7 ( subtract 8 from both sides )

a₁ = - 1

The sum to n terms of an arithmetic series is


S_(n) =
(n)/(2) [ 2a₁ + (n - 1)d ] , thus


S_(20) =
(20)/(2) [ (2 × - 1) + (19 × 4) ]

= 10(- 2 + 76) = 10 × 74 = 740

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