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# Question 71. Solve the equation (-4) + (-1) + 2 + 5 + …. + x = 437.

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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 the sum of arithmetic progression with a term x. And we have to find the value of x.

CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 71

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1. So, first term(a) = –4, common difference(d) = –1 – (–4) = 3, last term(an) = x and sum(Sn) = 437.

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

Sn = n[2a +(n – 1)d] / 2

We get

=> 437 = n[2(–4) + (n – 1)3] / 2

=> 874 = n[–8 + 3n – 3]

=> 3n– 11n – 874 = 0

=> 3n– 57n + 46n – 874 = 0

=> 3n(n – 19) + 46(n – 19) = 0

=> n = 19 or n = –46/3

Ignoring n = –46/3 as n cannot be a fraction as well as negative. So, we get n =19.

Now, we know nth term of an A.P. is given by an = a + (n – 1)d.

=>  x = –4 + (19 – 1)3

=> x = –4 + 54

=> x = 50

Hence, the value of x is 50.

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