Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]
⇒ a + (9-1)d = 0
⇒ a + 8d = 0 ⇒ a = -8d ...(i)
Now, its 19th term , T19 = a + (19-1)d
= - 8d + 18d [from Eq.(i)]
= 10d ...(ii)
and its 29th term, T29 = a+(29-1)d
= -8d + 28d [from Eq.(i)]
= 20d = 2 × T19
Hence, its 29th term is twice its 19th term
Answer:
Proved below.
Step-by-step explanation:
a9 = a1 + 8d = 0 where a1 = first term and d = common difference.
we need to prove that
a1 + 28d = 2(a1 + 18d
simplifying:-
a1 + 36d - 28d = 0
a1 + 8d = 0 which is what we are given.
Therefore the proposition is true.
Answer:
Step-by-step explanation:
For seven months it is less expensive to have a monthly membership.
When evaluating arithmetic expressions, the stack organization is extremely efficient. We employ a number of techniques, including PEMDAS, to simplify the issue. In addition to the value, the variables also need to be made easier.
Given, A full-year membership to a gym costs $325 upfront with no monthly charge. A monthly membership costs $100 upfront and $30 per month. For calculating the number of months is it less expensive to have a monthly membership we need to create an expression where the number of months will be variable. Hence,
100 + 30x < 325
subtract 100 on both sides
30x < 225
divide 30 into both sides
x < 225/30
x < 7.21
Therefore, it is less expensive to have a monthly membership if we take it for seven months.
Learn more about arithmetic expressions here:
#SPJ2
384,400
5,437,200,184,400
Part one: Arif and Keisha go to a restaurant. their meal cost 13.75. The sales tax is 5%
I got $0.69.
Part two: They want to leave a 15% tip based on the bill and tax combined. I got $2.17
Now the part 3 I need help with says Arif ordered a more expensive meal than Keisha after tax and tip are figured he decides he should pay $3.00 more than Keisha. How much should he pay?