The 9th and 22nd term of an A.P are 29 and 55 respectively,find the sum of its first 70 terms
Let the first term of the A.P be 'a' and the common difference be 'd'.
The formula for the nth term of an arithmetic progression is given by:
a_n = a + (n-1)d
Given that the 9th term is 29, we can use this information to set up an equation:
29 = a + 8d
Given that the 22nd term is 55, we can set up another equation:
55 = a + 21d
Now we have a system of equations with two variables. We can solve this system to find the values of 'a' and 'd'.
Multiplying the first equation by 3 and subtracting it from the second equation gives:
55 - 3(29) = a + 21d - 3(a + 8d)
55 - 87 = a + 21d - 3a - 24d
-32 = -2a - 3d
Rearranging this equation, we get:
2a + 3d = 32 ----(equation 3)
Now we can solve the system of equations consisting of equations 1 and 3.
Multiplying equation 1 by 2 and adding it to equation 3 gives:
2(29) + 2(8d) + 2a + 3d = 32 + 2a + 3d
58 + 16d = 32 + 2a + 3d
Simplifying this equation gives:
16d - 3d = 32 - 58
13d = -26
d = -2
Now we can substitute the value of 'd' back into equation 1 to find 'a':
29 = a + 8(-2)
29 = a - 16
a = 45
So the first term of the A.P is 45 and the common difference is -2.
To find the sum of the first 70 terms, we can use the formula for the sum of an arithmetic progression:
Sn = (n/2)(2a + (n-1)d)
Substituting the given values, we have:
S70 = (70/2)(2(45) + (70-1)(-2))
S70 = 35(90 + 69(-2))
S70 = 35(90 - 138)
S70 = 35(-48)
S70 = -1680
Therefore, the sum of the first 70 terms of the arithmetic progression is -1680.