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
-Added by ATP Admin Mathematics » Arithmetic Progressions
Answer:
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
= 1955 Answer
(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
= 109 Answer
(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
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