Class 10 Math All topics Arithmetic Progressions

In an A.P., the sum of the first 10 terms is 150, and the sum of the next 10 terms is 550. Find the A.P.

Solution:

From the question it is given that,

The sum of the first 10 terms $=−150$

The sum of the next 10 terms $=−550$

A.P $=?$

We know that, $S_{n}=(n/2)[2a+(n−1)d]$

$S_{10}=(n/2)[2a+(10−1)d]−150=(10/2)[2a+9d]−150=5[2a+9d]−150=10a+45d $

Then, $S_{20}=S_{10}+S_{10}$

$== −150−550−700 S_{20}=(20/2)[2a+19d]−700=10(2a+19d)−700=20a+190d $

Now, multiplying equation(i) by 2 we get,

$20a+90d=−300$

Subtract equation (iii) from equation (ii),

$(20a+190d)−(20a+90d)=−700−(−300)$

$20a+190d−20a−90d=−700+300100d=−400d=−400/100 $

$d=−4$

Substitute the value of d in equation (i) we get,

$10a+45(−4)=−15010a−180=−15010a=−150+18010a=30a=30/10a=3a_{2}=3+(−4)=3−4=−1a_{3}=−1+(−4)=−1−4=−5a_{3}=−5+(−4)=−5−4=−9 $

Therefore, A.P. is 3, -1, -5, -9, …

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