The 7th term of an arithmetic sequence is 40 and 13th term is 80.
(a) What is its 10th term ?
(b) What is the sum of first 19 terms of the sequence ?
(C) What is the sum of first 19 terms of the arithmetic sequence with 7h term 42 and
13 term 82 ?
Answers
Answer:
I only know answer for the first one....sorry.
Step-by-step explanation:
a₇ = a + 6d = 40 ...(1)
a₁₃ = a + 12d = 80 ...(2)
By subtracting (2) from (1), we get,
(1) ⇒ a + 6d = 40
(2)⇒ -a - 12d = -80
-6d = -40
d = -40/-6 = 6.67 ...(3)
By substituting (3) in (1), we get,
a + 6(6.67) = 40
= a = 0 ...(4)
a₁₀ = a + 9d ...(5)
By substituting (3) & (4) in (5), we get,
a₁₀ = 0 + 9(6.67)
= 60
∴ the 10th term is 60.
Answer:
(a) 60 (b) 506.667 (c) 392.667
Step-by-step explanation:
a₇=40 and a₁₃=80
we know that, a+(n-1)d=an
a+(7-1)d=40 , a+6d=40 and
a+(13-1)d=80, a+12d=80
from these two equations,
a=40-6d and a=80-12d
on equating, 40-6d=80-12d
6d=40
d=40/6=20/3
a=40-6(40/6) = 40-40=0
(a) a₁₀=a+(10-1)d=a+9d
=0 + 9(20/3)
a₁₀=60
(b)S₁₉= [19.{2a+(9-1)d}]/2
=[19.(2(0)+8(20/3))]/2
=[19.(160/3)]/2
=3040/6
S₁₉=506.667
(c)a₇=42 and a₁₃=82
we know that, a+(n-1)d=an
a+(7-1)d=42 , a+6d=42 and
a+(13-1)d=82, a+12d=82
from these two equations,
a=42-6d and a=82-12d
on equating, 42-6d=82-12d
6d=40
d=40/6=20/3
a=42-6(40/6) = 42-40=2
S₁₉=19/2.(2(2)+18(20/3))
=(19.(4+120))/6
=(19*124)/6
=2356/6
S₁₉=392.667