the sum of 3rd and 5th terms of an a.p is 38 and sum of 7th and 10th terms of an a.p is 83,find the a.p.
please help all questions please help please.
a+2d + a+4d=38 -->2a + 6d = 38
a+6d + a+9d = 83 --> 2a + 15d = 83
subtract them:
9d = 45
d=5
back in 2a+6d = 38
2a+30=38
a=4
the sequence is:
4 9 14 19 24 ...
BTW, the definitions used by Praphul are not correct.
e.g. he said:
5th term is (a+4)d
that should have been a + 4d, the brackets there are critical
To find the arithmetic progression (AP) given the sum of certain terms, we can use the formula for the nth term of an AP:
an = a + (n - 1)d
Where:
an = nth term of the AP
a = first term of the AP
d = common difference of the AP
n = position of the term in the AP
Now let's solve the problem step by step:
Step 1: Find the equations using the given information.
Given: The sum of the 3rd and 5th terms is 38.
We can write this as: (a + 2d) + (a + 4d) = 38
Given: The sum of the 7th and 10th terms is 83.
We can write this as: (a + 6d) + (a + 9d) = 83
Step 2: Simplify the equations and solve for 'a' and 'd'.
(a + 2d) + (a + 4d) = 38
2a + 6d = 38 (equation 1)
(a + 6d) + (a + 9d) = 83
2a + 15d = 83 (equation 2)
Step 3: Solve the set of equations to find the values of 'a' and 'd'.
To solve this system of equations, we can multiply equation 1 by 15 and equation 2 by 6 to eliminate the 'a' term.
15(2a + 6d) = 15(38)
6(2a + 15d) = 6(83)
30a + 90d = 570 (equation 3)
12a + 90d = 498 (equation 4)
Subtract equation 4 from equation 3 to eliminate 'd':
(30a + 90d) - (12a + 90d) = 570 - 498
18a = 72
a = 4
Substituting the value of 'a' into equation 1, we can solve for 'd':
2a + 6d = 38
2(4) + 6d = 38
8 + 6d = 38
6d = 38 - 8
6d = 30
d = 5
Therefore, the first term, 'a', is 4 and the common difference, 'd', is 5.
Step 4: Write the AP.
The arithmetic progression (AP) can be written as: 4, 9, 14, 19, 24, ...
Each term is found by adding 5 to the previous term.
To find the arithmetic progression (AP), we need to determine the common difference (d) and the first term (a).
Let's solve this step-by-step.
Step 1: Finding the common difference (d)
The difference between two consecutive terms in an AP is known as the common difference. We can use the given information to find the common difference.
Given: The sum of the 3rd and 5th terms of the AP is 38.
Let the 3rd term be a + 2d and the 5th term be a + 4d.
Therefore, (a + 2d) + (a + 4d) = 38
Simplifying the equation, we get:
2a + 6d = 38 --- (Equation 1)
Similarly,
Given: The sum of the 7th and 10th terms of the AP is 83.
Let the 7th term be a + 6d and the 10th term be a + 9d.
Therefore, (a + 6d) + (a + 9d) = 83
Simplifying the equation, we get:
2a + 15d = 83 --- (Equation 2)
Step 2: Solving the equations
We have two equations with two variables (a and d). We need to solve these equations simultaneously.
Let's simplify Equation 1 by dividing it by 2:
a + 3d = 19 --- (Equation 1')
Now, let's multiply Equation 1' by 3 and subtract it from Equation 2:
(2a + 15d) - 3(a + 3d) = 83 - (3 * 19)
2a + 15d - 3a - 9d = 83 - 57
-a + 6d = 26 --- (Equation 3)
Step 3: Solving Equation 3
To simplify the calculation, let's multiply Equation 3 by -1:
a - 6d = -26 --- (Equation 3')
Let's add Equation 3' and Equation 1':
(a - 6d) + (a + 3d) = -26 + 19
2a - 3d = -7 --- (Equation 4)
Step 4: Solving Equation 4
Let's solve Equation 4 using the substitution method.
Multiply Equation 3' by 2:
2(a - 6d) = 2(-26)
2a - 12d = -52
Now, add Equation 4 and the new equation we obtained:
(2a - 3d) + (2a - 12d) = -7 - 52
4a - 15d = -59 --- (Equation 5)
Step 5: Solving Equation 5
We now have a new equation with one variable, which we can easily solve.
Now, let's divide Equation 5 by 4:
4a - 15d = -59
a - (15d/4) = -(59/4) --- (Equation 5')
Step 6: Finding the value of a
From Equation 5', we know that a = (15d/4) - (59/4).
We can substitute this value of a into Equation 1':
(15d/4) - (59/4) + 3d = 19
Combining like terms, we get:
15d - 59 + 12d = 76
Simplifying the equation further:
27d = 135
d = 5
Step 7: Finding the value of a
We can substitute the value of d into Equation 1' to find the value of a:
a + 3(5) = 19
a + 15 = 19
a = 4
Therefore, the first term (a) is 4, and the common difference (d) is 5.
Step 8: Representing the arithmetic progression
Now that we know the first term (a = 4) and the common difference (d = 5), we can write the AP as follows:
AP = 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, ...
So, the arithmetic progression is 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, ...
We have the formula a+(n-1)d=nth term
Where a is the 1st term ,d is the difference and n is the nth term
So1st term= a
2nd term= (a+1)d
3rd term= (a+2)d
5th term=(a+4)d
Adding ad+2d+ad+ 4d= 38
2ad+ 6d= 38
In the same way 7th term=(a+6)d
10th term=(a+9)d
Adding these 2ad+15d=83
Subtracting, 9d= 45 which makes d=5