If third term of an A.P. is four times of first term and sixth term is 17. find the series.
first term -- a
term3 = a + 2d, but term3 = 4a
4a = a+2d
3a = 2d
term6 = a + 5d = 17
or a = 17-5d
sub that into 3a = 2d
3(17-5d) = 2d
51 - 15d = 2d
51 = 17d
d = 3
then 3a = 6
a = 2
terms are: 2, 5, 8, 11, 14, 17 ,
all checks out,
the fourth terms is 4 times the first
and the 6th is 17
Actually you need to find series that means sum of the sequence..
That is formula of Sn=?
By using Tn=Sn-S(n-1)
Thanks to solve this question with explanation
To find the series, we need to determine the value of the first term (a) and the common difference (d) in the arithmetic progression (A.P.).
Let's solve the problem step by step:
Step 1: Identify the given information.
- The third term of the A.P. is four times the first term.
- The sixth term of the A.P. is 17.
Step 2: Express the terms of the A.P. in terms of the first term (a) and the common difference (d).
The third term can be expressed as a + 2d since it is 2 terms after the first term (a).
The sixth term can be expressed as a + 5d since it is 5 terms after the first term (a).
Step 3: Set up the equations based on the given information.
According to the problem, we have:
a + 2d = 4a (From the first information: third term = 4 * first term)
a + 5d = 17 (From the second information: sixth term = 17)
Step 4: Solve the equations.
Using the first equation, we can simplify it by subtracting "a" from both sides:
2d = 3a
Now we can substitute this value into the second equation:
(3a) + 5d = 17
Simplify the equation by substituting 2d with 3a:
3a + 5(3a) = 17
3a + 15a = 17
18a = 17
a = 17/18
Step 5: Calculate the common difference (d).
Now that we know the value of the first term (a), we can substitute it back into the equation for the third term:
a + 2d = 4a
Substituting the value of "a" that we found:
(17/18) + 2d = 4(17/18)
Simplifying the equation:
17 + 36d = 68
36d = 68 - 17
36d = 51
d = 51/36
d = 17/12
Step 6: Write the series.
Now that we have found the values of the first term (a) and the common difference (d), we can write the series:
a, a + d, a + 2d, a + 3d, a + 4d, a + 5d
Substituting the values we found:
17/18, (17/18) + (17/12), (17/18) + 2(17/12), (17/18) + 3(17/12), (17/18) + 4(17/12), 17
Simplifying:
17/18, 17/18 + 17/12, 17/18 + 34/12, 17/18 + 51/12, 17/18 + 68/12, 17
Thus, the series is: 17/18, 17/18 + 17/12, 17/18 + 34/12, 17/18 + 51/12, 17/18 + 68/12, 17