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Question 1. Asked on :09 July 2019:07:05:00 PM

#### If the sum of nth terms of an A.P. is given by 4n2 - 7n. Find (i) Sum of 23 terms (ii) 15th term(iii) nth term

Master Mind

Given:Sn = 4n2 - 7n

Step by step explanation:

(i) Sum of 23 terms

Sn = 4n2 - 7n ............. (i

Replace n by 23 we have,

S23 = 4(23)2 - 7(23)

= 4 × 529 - 161

= 2116 - 161

(ii) 15th term

a15 = S15 - S14

S15 = 4(15)2 - 7(15) = 4 × 225 - 105 = 900 - 105 = 795

S14 = 4(14)2 - 7(14) = 4 × 196 - 98 = 784 - 98 = 686

Now, a15 = S15 - S14

= 795 - 686

(iii) nth term

an = Sn - Sn-1

Given, Sn = 4n2 - 7n

Replace n by n-1 we have

Sn-1 = 4(n-1)2 - 7(n-1)

= 4(n2 - 2n + 1) - 7n + 7

= 4n2 - 8n + 4 - 7n + 7

= 4n2 - 15n + 11 ............. (ii)

Now Applying equation (i) and (ii) in formula an = Sn - Sn-1

an = (4n2 - 7n) - (4n2 - 15n + 11)

= 4n2 - 7n - 4n2 + 15n - 11

= 8n - 11

Therefore, an = 8n - 11 Answer

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-Answered by Master Mind On 11 July 2019:09:42:19 PM(2123Average Rating Based on rating)

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