10th CBSE math solution for exercise 5.3 part 4

This page 10th CBSE math solution for exercise 5.3 part 4 is going to provide you solution for every problems that you find in the exercise no 5.3 part 4

10th CBSE math solution for exercise 5.3 part 4

(vi) Given a = 2 , d = 8,S n = 90 find n and a n

Solution:

S n = (n/2) [2a + (n - 1) d]

90 = (n/2) [2 (2) + (n - 1) 8]

 90 x 2 = n [4 + 8 n - 8]

 180 = n [-4 + 8 n]

  180 = -4 n  + 8 n²

   8 n² - 4 n - 180 = 0

   Divide the whole equation by 4

   2 n² - n - 45 = 0

(n - 5) (2 n + 9) = 0

n = 5   or n = -9/2

number of terms may not be negative or fraction.

a = a + 4 d

   = 2 + 4 (8) 

   = 2 + 32

   = 34


(vii) Given a = 8 , a n = 62 ,S n = 210 find n and d

Solution:

a n = a + (n - 1) d

a n = 8 + (n - 1) d

62 = 8 + (n - 1) d

62 - 8 = (n - 1) d

  54 = (n - 1) d -----(1)

S n = (n/2) [2a + (n - 1)d]

210 = (n/2) [ 2 (8) + (n - 1) d]

applying the first equation in this

210 = (n/2)[16 + 54]

 210 = (n/2) [70]

 210/70 = (n/2)

   3 x 2 = n

    n = 6

(n - 1) d = 54

 (6 - 1)d = 54

  5 d = 54

     d = 54/5


In the page 10th CBSE math solution for exercise 5.3 part 4 next we are going to see the solution of next problem.

(viii) Given an = 4 , d = 2 ,S n = -14 find n and a

Solution:

a n = a + (n - 1) d

a n = a + (n - 1) (2)

4 = a + 2 n - 2

4 + 2 = a + 2 n

6 = a + 2 n

 a + 2 n = 6 ------(1)

a n = l = 4

 S n = (n/2) [a + l]

-14 = (n/2) [ a + 4]

(-14 x 2) = n [ a + 4]

-28 = n[6- 2 n + 4]

-28 = n [10 - 2 n]

-28 = 10 n - 2 n²

2 n² - 10 n - 28 = 0

  n² -  5 n - 14 = 0

 (n - 7) (n + 2) = 0

 n = 7  or n = -2

Substitute n = 7 in the first equation

 a + 2(7) = 6

 a + 14 = 6

  a = 6 - 14

 a = -8




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