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26) Find the middle terms of the sequence formed by all numbers between 9 and 95, which leavea remainder 1 when divided by 3. Also, find the sum of the numbers on both sides ofmiddle term separately |
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Answer» The sequence are formed by all the numbers between 9 and 95 , which leaves remainder 1 when it is divided by 3 : 10, 13, 16, .........94 Here you can see 10, 13 , 16 , .....94 are in AP where first term , a = 10 and common difference , d = 94 . From AP nth term formula , Tn = a + (n -1)d 94 = 10 + (n -1)3 84/3 = n -1 ⇒n = 29 Hence, there are 29 terms between 10 and 94 It means middle term is (n + 1)/2 th term e.g., middle term = (29+1)/2 = 15th term So, T₁₅ = a + (15 - 1)d = 10 + 14 ×3 = 52 Hence, 15th term is 52 here middle is 15th , it means 14 terms are in left side of it and 14 terms are in right side of it. So, sum of first 14 term {left side } = 14/2 [ 2 × 10 + (14 -1)×3 ] [∵Sn = n/2[2a + (n-1)d]= 7[20 + 39] = 7 × 59 = 413 Now, sum of last 14 term { right side of middle term } = S₂₉ -[ S₁₄ + T₁₅ ]= 29/2[2 × 10 + (29-1) × 3 ] - 413 - 52 = 29/2[20 + 84 ] - 465 = 1508 - 465 = 1043 |
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