In an A.P., the sum of first ten terms is −150

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.

Solution:

Here, we are given  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 10th term (n = 10),

$a_{10}=a+(10-1) d$

$=a+9 d$

So, here we can find the sum of the $n$ terms of the given A.P., using the formula, $S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P,

$S_{10}=\left(\frac{10}{2}\right)(a+a+9 d)$

$-150=5(2 a+9 d)$

$-150=10 a+45 d$

$a=\frac{-150-45 d}{10}$ ..(1)

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

First term = a11

Last term = a20

For the 11th term (n = 11),

$a_{11}=a+(11-1) d$

$=a+10 d$

For the 20th term (n = 20),

$a_{20}=a+(20-1) d$

$=a+19 d$

So, for the given A.P,

$S_{10}=\left(\frac{10}{2}\right)(a+10 d+a+19 d)$

$-550=5(2 a+29 d)$

$-550=10 a+145 d$

$a=\frac{-550-145 d}{10}$ ..........(2)

Now, subtracting (1) from (2),

$a-a=\left(\frac{-550-145 d}{10}\right)-\left(\frac{-150-45 d}{10}\right)$

$0=\frac{-550-145 d+150+45 d}{10}$

$0=-400-100 d$

$100 d=-400$

$d=-4$

Substituting the value of in (1)

$a=\frac{-150-45(-4)}{10}$

$=\frac{-150+180}{10}$

$=\frac{30}{10}$

 

$=3$

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

 

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