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

(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

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