Learn all Concepts of Chapter 5 Class 10 (with VIDEOS). Check - Arithmetic Progressions - Class 10     1. Chapter 5 Class 10 Arithmetic Progressions
2. Serial order wise
3. Ex 5.4 (Optional)

Transcript

Ex 5.4, 2 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, 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 Product of the third and seventh terms is 8 a3 × a7 = 8 (a + 2d) (a + 6d) = 8 a (a + 6d) + 2d (a + 6d) = 8 a2 + 6ad + 2ad + 12d2 a2 + 8ad + 12d2 = 8 We factorize by splitting the middle term a2 + 2ad + 6ad + 12d2 = 8 a(a + 2d) + 6d (a + 2d) = 8 (a + 2d) (a + 6d) = 8 Hence, our equations are a + 4d = 3 …(1) (a + 2d) (a + 6d) = 8 …(2) From equation (1) a + 4d = 3 a = 3 − 4d Putting a = 3 − 4d in equation (2) (a + 2d) (a + 6d) = 8 (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 Taking d = 1/2 Put in equation (1) a + 4d = 3 a + 4 (1/2) = 3 a + 2 = 3 a = 1 Take d = (−1)/2 Put in equation (1) a + 4d = 3 a + 4 ((−1)/2) = 3 a − 2 = 3 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 S16 = 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 S16 = 20 Hence, if a = 1 and d = 𝟏/𝟐 , the sum of first sixteen terms o f the AP is 76 and if a = 5 and d = (−𝟏)/𝟐, the sum of first sixteen terms of the AP is 20.

Ex 5.4 (Optional) 