what is the difference between the 11th and 17th term of an arithmetic series having a common difference of 9
Answers
Answer:
Given: 6th term A6= 20, 11th term A11=30
we have, An= A1+(n-1)d
=> A6=A1+(6–1)d
=> 20=A1+5d ——equation (1)
A11=A1+(11–1)d
=> 30=A1+(11–1)d
=> 30= A1+10d ——equation (2)
equating 1 and 2, we have
=> d=2, substitute in eq(1)
=> 20= A1+5(2)
=>A1=20–10=10
Hence, the first term A1=10 and common difference d=2.
Step-by-step explanation:
Given: 6th term A6= 20, 11th term A11=30
we have, An= A1+(n-1)d
=> A6=A1+(6–1)d
=> 20=A1+5d ——equation (1)
A11=A1+(11–1)d
=> 30=A1+(11–1)d
=> 30= A1+10d ——equation (2)
equating 1 and 2, we have
=> d=2, substitute in eq(1)
=> 20= A1+5(2)
=>A1=20–10=10
Hence, the first term A1=10 and common difference d=2.
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The 1st term and the 11th term of an arithmetic progression are 5 and 45 respectively. What is the common difference of the progression?
The 7th and 12th terms of an arithmetic progression are -15 and 5 respectively. What is the 16th term?
They look like this, a, a+d, a+2d.. a + (n-1)d
6th and 11th are 20 and 30
The 6th term is a+5d while the 11th is a + 10d
a+10d = 30
a+5d = 20, subtract these two
5d = 10, d=2
Plug this d back in to get a+20=30, a = 10
A is the first term while d is the common difference so
the first term is 10 and the common difference is 2.
The easiest way to work this problem is to note that you have to add five common differences to get from the 6th term to the 11th term. So the common difference is simply (30–20)/5 or 2.
Now to find the 1st term, note that it is five common differences away from the 6th term and that 5 common distances is 10.
So the first term is 20 - 10 = 10.
nth term of AP =a+(n-1)d
6th term of AP=20=a+5d ………….(1)
11 the term of AP=30=a+10d……….(2)
Solving (1) and (2) we get
a=10 and d=2
T6 : a+5d = 20 …(1)
T11 : a+10d = 30 …(2)
Subtract (1) from (2)
5d = 10, or d = 2.
So a, the first term = 20–10 = 10.
The first term is 10 and the common difference is 2.
6th term =a+5d=20…eq1
11th term =a+10d=30….eq2
On solving simultaneously we get
d=2
Term 1=a
a=10
Difference between 6th term & 11th term is 10.
=> Common difference = 10/5 = 2
First term = 20 - ( 2 x 5) = 10
:-)
The present age of the father and the son are 34 years and 4 years. How many years ago was the product of their age was 31 years?
If x is the number of years ago, then the equation to solve is:
(34-x)(4-x)=31
Since 31 is prime, it kind of jumps out immediately that the answer is three years ago, when the father was 31 and the son was one. But that’s not helpful, let’s pretend we didn’t notice and solve it “the real way”.
Continuing the math:
Expand with FOIL:
136–34x-4x+x^2 = 31
Combine like terms:
x^2–38x+105 = 0
Factor:
The digits in 105 add up to 6, which is a multiple of 3, so 105 is divisible by 3. Try dividing by 3, you get 35. Hey, 3+35=38, so we’re in business.
(x-3)(x-35)=0
So x is either 3 or 35.
3 is obviously the real an
Which term of the progression 20, 19¼, 18½, 17¾, … is the first negative term?
The above sequence is arithmatic progression
The first term of the sequence is 20 i.e
a=20
The common difference is of -3/4
We can calculate common difference by subtracting first term from second term i.e a2-a1
So now we have first term as well as common difference then we can calculate the n th term
To calculate n th term we will use formula
Tn=a+(n-1)d
Where a=first term
d=common difference
So by putting this values of a and d in the above equation we get
Tn=20+(n-1)(-3/4)
First we put n=1 and we get the second term which is 19 1/4
Now also the common difference is negative and also it is less than 1 we
Ten years ago, the ratio of the ages of a woman to her daughter was 3:2. Which of the following cannot be the ratio of their ages 5 years from now? (a) 6:5 (b) 7:3 (c) 8:7 (d) 11:9
Let say 10 year ago mother’s age was 3c and daughter’s age was 2c.
Now at particular time t, their ages will be :
y1 = t + 3c
y2 = t + 2c { assuming 10 year ago t=0}
Both are the equstions of line where x axis is time and y axis is age. Both lines are parallel to each other.
As t tends to infinite y1/y2 tends to 1.
So, min value of y1/y2 will be 1 and max value wiĺl be 1.5 ( at t = 0 ).
So, the range of y1/y2 will be (1, 1.5].
Option (b) 7:3 is out of range. Option b is the correct answer.
What is the smallest number, which, when divided by 12, 15, 18, and 27, leaves remainder of 8, 11, 14 and 23, respectively?
Prime factorizations first.
12 = 2 x 2 x 3
15 = 3 x 5
18 = 2 x 3 x 3
27 = 3 x 3 x 3
LCM = 2 x 2 x 3 x 3 x 3 x 5 = 540
Since all of the remainders are 4 less than the divisors, the number will be 540 - 4 = 536
The 3rd and 7th term of an arithmetic ogression are 6 and 0 respectively. What are the common difference, first term and 10th term?
Let a and d be the first term and the common difference respectively. Since the 3rd and the 7th terms of the AP are 6 and 10 respectively, then:
a +2d = 6………..(i) and
a + 6d = 0………(ii).
From (ii),
a = -6d…..(iii).
Substituting -6d for a in (i), we have :
-6d + 2d = 6.
This follows that:
-4d = 6,
which follows that:
one of the numbers is 24 the other has to be 36 . Their difference follows immediately.