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
1 answer:
Answer:
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
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