24. Prove that tangent at any ponit 0l acicnum of first six terms of a

1.	24. Prove that tangent at any ponit 0l acicnum of first six terms of an AP is 42. The ratio of its 10th term to 30th term is 1:3. Calculatethfirst and 13th term.A60
Answer» Sum of first 6 terms of AP = 42S6 = 6/2(2a + 5d)...... (1) nth term of an APTn = a + (n-1)dWhere, a = first termd = common difference According to the given condition T10/T30 = 1/3(a + 9d)3 = a + 29d3a + 27d = a + 29d2a = 2da = d Put value of d = a in eq(1)42 = 3(2a + 5a)7a = 14a = 14/7 = 2 Thus a = 2, d = 2 Therefore, First term = 213th term = a + 12d = 2 + 12*2 = 26

24. Prove that tangent at any ponit 0l acicnum of first six terms of an AP is 42. The ratio of its 10th term to 30th term is 1:3. Calculatethfirst and 13th term.A60

Answer»

Sum of first 6 terms of AP = 42S6 = 6/2(2a + 5d)...... (1)

nth term of an APTn = a + (n-1)dWhere, a = first termd = common difference

According to the given condition T10/T30 = 1/3(a + 9d)3 = a + 29d3a + 27d = a + 29d2a = 2da = d

Put value of d = a in eq(1)42 = 3(2a + 5a)7a = 14a = 14/7 = 2

Thus a = 2, d = 2

Therefore, First term = 213th term = a + 12d = 2 + 12*2 = 26