The 9th and 22nd term of an A.P are 29 and 55 respectively,find the sum of its first 70 terms

Answer 1

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.