Ex 5.4, 2 (Optional) - The sum of third and seventh terms of AP is 6

Ex 5.4, 2 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.4, 2 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 3
Ex 5.4, 2 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 4
Ex 5.4, 2 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 5

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Ex 5.4, 2 (Optional) The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. We know that nth term of an AP is an = a + (n − 1)d Hence, 3rd term of AP = a3 = a + 2d and 7th term of AP = a7 = a + 6d Given Sum of third & seventh terms is 6 a3 + a7 = 6 a + 2d + a + 6d = 6 2a + 8d = 6 2(a + 4d) = 6 a + 4d = 6/2 a + 4d = 3 a3 + a7 = 6 a + 2d + a + 6d = 6 2a + 8d = 6 2(a + 4d) = 6 a + 4d = 6/2 a + 4d = 3 Also, Product of the third and seventh terms is 8 a3 × a7 = 8 (a + 2d) (a + 6d) = 8 From (1) a + 4d = 3 a = 3 − 4d (3 − 4d + 2d) (3 − 4d + 6d) = 8 (3 − 2d) (3 + 2d) = 8 (3)2 − (2d)2 = 8 9 – 4d2 = 8 4d2 = 1 (2d)2 = (1)2 2d = ± 1 d = ± 𝟏/𝟐 Finding value of a For d = 𝟏/𝟐 a = 3 − 4d a = 3 − 4(1/2) a = 3 − 2 a = 1 For d = (−𝟏)/𝟐 a = 3 − 4d a = 3 − 4((−1)/2) a = 3 + 2 a = 5 For d = (−𝟏)/𝟐 a = 3 − 4d a = 3 − 4((−1)/2) a = 3 + 2 a = 5 Therefore, when a = 1, d = 𝟏/𝟐 And when a = 5, d = (−𝟏)/𝟐 Now, we need to find the Sum of first Sixteen Terms Sum of n terms of an AP is Sn = 𝒏/𝟐 [𝟐𝒂+(𝒏 −𝟏)𝒅] Taking a = 1 and d = 𝟏/𝟐 S16 = 16/2 [(2" × 1" )+(16 −1)(1/2)] = 8 [2+15/2] = 8 [(4 + 15)/2] = 8 × 19/2 = 76 Taking a = 5 and d = (−𝟏)/𝟐 S16 = 16/2 [(2" × " 5)+(16 −1)((−1)/2)] = 8 [10−15/2] = 8 [(20 − 15)/2] = 8 × 5/2 = 20 Hence, If a = 1 and d = 𝟏/𝟐 , the sum of first sixteen terms of the AP is 76 and If a = 5 and d = (−𝟏)/𝟐, the sum of first sixteen terms of the AP is 20.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.