Question

In an AP, the sum of the first ten terms is -150 and the sum of its next ten terms is -550. Find AP.

Answer 1

Hint: Sum of n terms of AP ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$.

Given sum of first ten terms = $ - 150$
We know that Sum of n terms of AP ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
Therefore,
        $\
   \Rightarrow {S_{10}} = \dfrac{{10}}{2}[2a + (10 - 1)d] \\
   \Rightarrow - 150 = 5[2a + 9d] \\
   \Rightarrow - 150 = 10a + 45d \to (1) \\
\ $
And also given that sum of next ten terms =$ - 550$ (which also includes sum of first ten terms value which has to be removed)
Sum of next ten terms = $ - 550$
 $ \Rightarrow $$ - 150$$ - 550$ = $\dfrac{{20}}{2}[2a + (20 - 1)d]$
$\
   \Rightarrow - 700 = 10(2a + 19d) \\
   \Rightarrow - 70 = 2a + 19d \to (2) \\
\ $
Multiplying equation $(2) \times 5$ then we get
$ \Rightarrow - 350 = 10a + 95d \to (3)$
On solving $(1)\& (3)$ we get
$d = - 4$
Now by substituting $'d'$ value in equation $(1)$ we get
$\
   \Rightarrow - 150 = 10a + 45d \\
   \Rightarrow - 150 = 10a + 45( - 4) \\
   \Rightarrow 10a = - 150 + 180 \\
   \Rightarrow 10a = 30 \\
   \Rightarrow a = 3 \\
\ $
 Hence we got the value $a = 3,d = - 4$
We know that for an AP series $'a'$ be the first term and $'d'$ is the difference between the terms.
We also know that AP series will be of the form $a,a + d,a + 2d,a + 3d.......$
On substituting the $'a'$ and $'d'$ values
We get the values of series as 3,-1,-5,-9
Then the AP series will be $3, - 1, - 5, - 9....$

Note: In the above problem second condition i.e. sum of next ten terms includes sum of first 10 terms plus the other ten terms (where sum of first ten terms need to be subtracted from sum given for second condition) .Ignoring such simple condition will affect the answer.