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# Question 33. In an A.P. the sum of first ten terms is −150 and the sum of its next 10 term is −550. Find the A.P.

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One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.6
This type of question has been asked in previous years exams.

In this question we have been given that In an A.P. the sum of first ten terms is −150 and the sum of its next 10 term is −550.

Now we have to find the A.P.

CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 33

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1. Sum of first ten terms, S10 = −150.

Also given, Sum of its next 10 terms = −550

Sum of first 20 terms, S20 = Sum of first 10 terms + Sum of next 10 terms

=> S20 = −150 + (−550) = −700

By using the formula of the sum of n terms of an A.P.

Sn = n[2a + (n − 1)d] / 2.

So, S10 = 10(2a + (10 − 1)d) / 2

=> −150 = 5(2a + 9d)

=> 2a + 9d = –30  ….. (1)

And also, S20 = 20(2a + (20 − 1)d) / 2

=> −700 = 10(2a + 19d)

=> 2a + 19d = −70  ….. (2)

Now, subtracting eq(1) from (2), we get

=> 19d – 9d = –70 – (–30)

=> 10d = –40

=> d = –4

On putting d = –4 in (1), we get,

=> 2a + 9(–4) = –30

=> 2a = 6

=> a = 3

As we have a = 3 and d = –4, hence, the A.P is 3, –1, –5, –9,…..

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