The sum of the 1st and 2nd terms of an AP is 4 and the 10th term is 19. Find the sum of the 5th and 6th terms
a + a+d = 4
a+9d = 19
so, find a and d, and then you want
a+4d + a+5d
let 'a' is the first term of AP. 'd' is a common difference of AP.
and the AP is : a,a+d,a+2d..........
according to ques.
a1+a2=4
a+a+d=4......................................(1)
a10=19
a+9d=19......................................(2)
after solving the equation 1 and 2 we get the value of 'a' and 'd' is 1 and 2 respectively.
according to question we need to find the value of a5+a6=?
a+4d+a+5d=2a+9d ..............(3)
after puttig the value 'a' and 'd' in the 3 equation
we get, 2a+9d = 20
2a + d = 4
a + 9d =19. .
Solve simultaneously ,a=1 .,d =2. Find 2a + 9d =2 +18 =20
let 'a' is the first term of AP. 'd' is a common difference of AP.
and the AP is : a,a+d,a+2d..........
according to ques.
a1+a2=4
a+a+d=4......................................(1)
a10=19
a+9d=19......................................(2)
after solving the equation 1 and 2 we get the value of 'a' and 'd' is 1 and 2 respectively.
according to question we need to find the value of a5+a6=?
a+4d+a+5d=2a+9d ..............(3)
after puttig the value 'a' and 'd' in the 3 equation
we get, 2a+9d = 20
present
Why did the math book go to the comedy club?
Because it wanted to find the sum of the 5th and 6th terms of an arithmetic progression!
To find the sum of the 5th and 6th terms, we first need to find the common difference (d) of the arithmetic progression.
We know that the 10th term (a10) is 19, so we can use the formula: a10 = a1 + (10-1)d
Substituting the given values, we get:
19 = a1 + 9d
We also know that the sum of the 1st and 2nd terms (a1 + a2) is 4.
So, we can write:
a1 + a2 = 4
Now, we have two equations:
19 = a1 + 9d
a1 + a2 = 4
To find the sum of the 5th and 6th terms (a5 + a6), we can use the formula for the nth term of an arithmetic progression: an = a1 + (n-1)d
Substituting n = 5 and n = 6, we get:
a5 = a1 + 4d
a6 = a1 + 5d
Adding these two equations, we get:
(a5 + a6) = 2a1 + 9d
But we still have two unknowns, a1 and d. We need to find a way to eliminate one of them.
From the equation a1 + a2 = 4, we can solve for a1 in terms of a2:
a1 = 4 - a2
Substituting this into the equation (a5 + a6) = 2a1 + 9d, we get:
(a5 + a6) = 2(4 - a2) + 9d
(a5 + a6) = 8 - 2a2 + 9d
Now, let's solve for a2 in terms of d using the equation 19 = a1 + 9d:
a2 = 19 - a1 - 9d
Substituting this into the equation above, we get:
(a5 + a6) = 8 - 2(19 - a1 - 9d) + 9d
(a5 + a6) = 8 - 38 + 2a1 + 18d + 9d
(a5 + a6) = 2a1 - 11d - 30
Since we don't have the explicit values of a1 and d, we can't calculate the exact sum of the 5th and 6th terms. I apologize for the confusion. But hey, at least we had a good math joke, right?
To find the sum of the 5th and 6th terms of an arithmetic progression (AP), we first need to find the common difference (d) of the AP.
Given that the sum of the 1st and 2nd terms of the AP is 4, we can write the equation:
(a + (a+d)) = 4, where a is the first term.
Simplifying this equation, we get:
2a + d = 4 ----(Equation 1)
We are also given that the 10th term of the AP is 19. In general, the nth term (Tn) of an arithmetic progression can be calculated using the formula:
Tn = a + (n-1)d, where a is the first term, n is the term number, and d is the common difference.
Using this formula, we can write the equation for the 10th term:
a + (10-1)d = 19
Simplifying this equation, we get:
a + 9d = 19 ----(Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) with two variables (a and d). We can solve this system of equations to find the values of a and d.
Step 1: Multiply Equation 1 by 9:
18a + 9d = 36 ----(Equation 3)
Step 2: Subtract Equation 3 from Equation 2:
(a + 9d) - (18a + 9d) = 19 - 36
-17a = -17
Step 3: Divide both sides by -17:
a = 1
Substituting the value of a back into Equation 1, we can find the value of d:
2(1) + d = 4
2 + d = 4
d = 2
So, the first term (a) of the AP is 1, and the common difference (d) is 2.
Now that we have the first term (a = 1) and the common difference (d = 2), we can find the 5th and 6th terms of the AP:
T5 = a + (5-1)d = 1 + 4(2) = 1 + 8 = 9
T6 = a + (6-1)d = 1 + 5(2) = 1 + 10 = 11
Finally, we can find the sum of the 5th and 6th terms:
T5 + T6 = 9 + 11 = 20
Therefore, the sum of the 5th and 6th terms of the given AP is 20.