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Find the sum given below: –5 + (–8) + (–11) + ... + (–230)

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Question

Find the sum given below:

–5 + (–8) + (–11) + ... + (–230)

Sum
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Solution

–5 + (–8) + (–11) + ... + (–230)

For this A.P.,

a = −5

l = −230

d = a2 − a1 

= (−8) − (−5)

= − 8 + 5

= −3

Let −230 be the nth term of this A.P.

l = a + (n − 1)d

−230 = − 5 + (n − 1) (−3)

−225 = (n − 1) (−3)

(n − 1) = 75

n = 76

And Sn = `n/2(a+1)`

= `76/2[(-5)+(-230)]`

= 38 × (-235)

= -8930

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Chapter 5: Arithmetic Progressions - EXERCISE 5.3 [Page 68]

APPEARS IN

NCERT Mathematics [English] Class 10
Chapter 5 Arithmetic Progressions
EXERCISE 5.3 | Q 2. (iii) | Page 68
ML Aggarwal Understanding Mathematics [English] Class 10 ICSE
Chapter 9 Arithmetic and Geometric Progressions
Exercise 9.3 | Q 3.2
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