# Find the sum: 1 + (–2) + (–5) + (–8) + ... + (–236) - Mathematics

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Sum

Find the sum: 1 + (–2) + (–5) + (–8) + ... + (–236)

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#### Solution

Here, first term (a) = 1

And common difference (d) = (–2) – 1 = –3

∵ Sum of n terms of an AP

S_n = n/2 [2a + (n - 1)d]

⇒ S_n = n/2 [2 xx 1 + (n - 1) xx (-3)]

⇒ S_n = n/2 (2 - 3n + 3)

⇒ S_n = n/2 (5 - 3n)   .....(i)

We know that

If the last term (l) of an AP is known, then

l = a + (n – 1)d

⇒ – 236 = 1 + (n – 1)(– 3) ......[∵ l = – 236, given]

⇒ – 237 = – (n – 1) × 3

⇒ n – 1 = 79

⇒ n = 80

Now, put the value of n in equation (i), we get on

S_n = 80/2 [5 - 3 xx 80]

= 40[5 - 240]

= 40 xx (-235)

= – 9400

Here, the required sum is – 9400

Concept: Sum of First n Terms of an A.P.
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.3 | Q 21.(i) | Page 53
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