Question

In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.

Answer 1

Here, we are given  S₁₀ = -150  and sum of the next ten terms is −550.

Let us take the first term of the A.P. as a and the common difference as d.

So, let us first find a₁₀. For the sum of first 10 terms of this A.P,

First term = a

Last term = a₁₀

So, we know,

a_n = a + (n-1)d

For the 10^th term (n = 10),

a₁₀ = a + (10 -1)d 

      = a + 9d

So, here we can find the sum of the n terms of the given A.P., using the formula,  `S_n = (n/2) (a + l) `

Where, a = the first term

l = the last term

So, for the given A.P,

`S_10 = (10/2 ) (a +a + 9d)`

-150 = 5 (2a + 9d)

-150 = 10a  + 45 d

   `a = (-150-45d)/10`                           ................(1)

Similarly, for the sum of next 10 terms (S₁₀),

First term = a₁₁

Last term = a₂₀

For the 11^th term (n = 11),

a₁₁ = a + (11 - 1) d

      = a + 19d

For the 20^th term (n = 20),

a₂₀ = a + (20 - 1) d

      = a + 19d

So, for the given A.P,

`S_10 = (10/2) ( a + 10d + a + 19 d)`

-550 = 5(2a + 29d)

-550 = 10a + 145d

   `a = (-550-145d)/10`                   ...............(2) 

Now, subtracting (1) from (2),

    a-a=`((-550-145d)/10) - ((-150-45d)/10)`

     `0 = (-550-145d+150+45d)/10`

      0 = -400 -100d

100d = -400

     d = -4 

Substituting the value of d in (1)

`a = (-150-45(-4))/10`

   `=(-150+180)/10`

   `=30/10`

    = 3

So, the A.P. is 3,-1,-5,-9,... with a=3,d=-4 .