10th CBSE math solution for exercise 5.3 part 8
This page 10th CBSE math solution for exercise 5.3 part 8 is going to provide you solution for every problems that you find in the exercise no 5.3 part 8
10th CBSE math solution for exercise 5.3 part 8
(11) If the sum of the first n terms of an AP is 4 n - n² what is the first term (that is S)? what is the sum of first two terms?what is the second term? similarly find the 3rd,the 10th and the nth terms.
Solution:
S n = 4 n - n²
n = 1
= 4 (1) - (1)²
= 4 - 1 = 3
first term = 3
S n = 4 n - n²
n = 2
= 4 (2) - (2)²
= 8 - 4 = 4
Sum of first two terms = 4
to find the second term,we have to subtract the sum of two terms from first term
= 4 - 3
= 1
second term = 1
a ₃ = a + 2 d
d = second term - first term
= 1 - 3 = -2
a ₃ = 3 + 2 (-2)
= 3 - 4 = -1
a₁₀= a + 9 d
= 3 + 9(-2)
= 3 - 18
= -15
(12) Find the sum of first 40 positive integers divisible by 6.
Solution:
The positive integers which are divisible by 6 are
6,12,18,...........
6 + 12 + 18 + ...........
a = 6 d = 12 - 6 n = 40
= 6
Sn= (n/2) [ 2 a + (n - 1) d]
S₄₀= (40/2) [2(6) + (40-1) (6)]
= 20 [12 + 39(6)]
= 20 [12 + 234]
= 20 [246]
= 4920
In the page 10th CBSE math solution for exercise 5.3 part 8 next we are going to see the solution of next problem.
(13) Find the sum of first 15 multiples of 8.
Solution:
first 15 multiples of 8 are
8,16,24,................
8 + 16 + 24 + ...........
a = 8 d = 16 - 8 n = 15
= 8
Sn= (n/2) [ 2 a + (n - 1) d]
S₁₅ = (15/2) [2(8) + (15-1) (8)]
= (15/2) [16 + 14(8)]
= (15/2) [16 + 112]
= (15/2) [128]
= 15 [64]
= 960
