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Which term of the AP: –2, –7, –12,... will be –77? Find the sum of this AP upto the term –77.

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Question

Which term of the AP: –2, –7, –12,... will be –77? Find the sum of this AP upto the term –77.

Sum
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Solution

Given, AP: –2, –7, –12,...

Let the nth term of an AP is –77

Then, first term (a) = –2 and

Common difference (d) = –7 – (–2)

= –7 + 2

= –5

∵ nth term of an AP, Tn = a + (n – 1)d

⇒ –77 = –2 + (n – 1)(–5)

⇒ –75 = –(n – 1) × 5

⇒ (n – 1) = 15

⇒ n = 16

So, the 16th term of the given AP will be –77

Now, the sum of n terms of an AP is

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

So, sum of 16 terms i.e., upto the term –77

S16 = `16/2 [2 xx (-2) + (n - 1)(-5)]`

= 8[–4 + (16 – 1)(–5)]

= 8(–4 – 75)

= 8 × (–79)

= –632

Hence, the sum of this AP upto the term –77 is –632.

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Chapter 5: Arithematic Progressions - Exercise 5.3 [Page 53]

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NCERT Exemplar Mathematics [English] Class 10
Chapter 5 Arithematic Progressions
Exercise 5.3 | Q 22 | Page 53

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