Answers: (i)  n = 10 First term, a = a1 = 50 Common difference, d = a2 – a1 = 46 – 50 = -4 By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values of ‘a’ and ‘d’, we get S = 10/2 (100 +...

Answers: (i)  n = 25 First term, a = a1 = 3 Common difference, d = a2 – a1 = 9/2 – 3 = (9 – 6)/2 = 3/2 By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values of ‘a’ and ‘d’, we get S =...

Answers: (i)  n = 22 First term, a = a1 = a+b Common difference, d = a2 – a1 = (a-b) – (a+b) = a-b-a-b = -2b By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values of ‘a’ and ‘d’, we...

Answer: n = n First term, a = a1 = (x-y)/(x+y) Common difference, d = a2 – a1 = (3x – 2y)/(x + y) – (x-y)/(x+y) = (2x – y)/(x+y) By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values...

Answers: (i) First term, a = a1 = 2 Common difference, d = a2 – a1 = 5 – 2 = 3 an term of given AP is 182 an = a + (n-1) d 182 = 2 + (n-1) 3 182 = 2 + 3n – 3 182 = 3n – 1 3n = 182 + 1 n = 183/3 n =...

Answer: First term, a = a1 = (a-b)2 Common difference, d = a2 – a1 = (a2 + b2) – (a – b)2 = 2ab an term of given AP is [(a + b)2 + 6ab] an = a + (n-1) d [(a + b)2 + 6ab] = (a-b)2 + (n-1)2ab  ...

Answer: First term, a = a1 = 1 Common difference, d = a2 – a1 = 2 – 1 = 1 l = n Sum of n terms = S S = n/2 [2a + (n-1) d] S = n/2 [2(1) + (n-1) 1] S = n/2 [2 + n – 1] S = n/2 [n – 1] ∴ The sum of...

Answer: The natural numbers which are divisible by 2 or 5 are: 2 + 4 + 5 + 6 + 8 + 10 + … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) are AP with...

Answer: Given, AP of first n odd natural numbers whose first term a is 1 Common difference d = 3 The sequence is 1, 3, 5, 7……n a = 1, d = 3-1 = 2, n = n By using the formula, S = n/2 [2a + (n-1)d] S...

Answer: The odd numbers between 1 and 1000 divisible by 3 are 3, 9, 15,…,999 Number of terms be ‘n’, so the nth term is 999 a = 3, d = 9-3 = 6, an = 999 an = a + (n-1)d 999 = 3 + (n-1)6 999 = 3 + 6n...

Answer: The series is 85, 90, 95, …, 715 ‘n’ terms in the AP a = 85, d = 90-85 = 5, an = 715 an = a + (n-1)d 715 = 85 + (n-1)5 715 = 85 + 5n – 5 5n = 715 – 85 + 5 5n = 635 n = 635/5 n = 127 By using...

Answer: The series of integers divisible by 7 between 50 and 500 are 56, 63, 70, …, 497 Number of terms be ‘n’ a = 56, d = 63-56 = 7, an = 497 an = a + (n-1)d 497 = 56 + (n-1)7 497 = 56 + 7n – 7 7n...

Answer: All even integers will have a common difference of 2. AP is 102, 104, 106, …, 998 a = 102, d = 104 – 102 = 2, an = 998 By using the formula, an = a + (n-1)d 998 = 102 + (n-1)2 998 = 102 + 2n...