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

740

Step-by-step explanation:

The n th term of an arithmetic series is

= 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

= [ 2a₁ + (n - 1)d ] , thus

= [ (2 × - 1) + (19 × 4) ]

     = 10(- 2 + 76) = 10 × 74 = 740

answer: 1, 3 and 5 are correct

step-by-step explanation: