This page 10th CBSE math solution for exercise 5.2 part 5 is going to provide you solution for every problems that you find in the exercise no 5.2 part 5

10th CBSE math solution for exercise 5.2 part 5

(7) Find the 31st term of an AP whose 11th term is 38 and 16th term is 73

Solution:

a₁₁ = 38

a + 10 d = 38 ----- (1)

a₁₆ = 73

a + 15 d = 73 ----- (2)

(1) - (2)

a + 10 d = 38

a + 15 d = 73

(-) (-) (-)

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- 5 d = -35

d = 35/5

d = 7

Substitute d = 7 in the first equation

a + 10 (7) = 38

a + 70 = 38

a = 38 - 70

a = -32

a₃₁ = a + 30 d

= -32 + 30 (7)

= -32 + 210

= 178

(8) An AP consists of 50 terms of which 3rd term is 12 and the last term is 73.

Solution:

a₃ = 12

l = 73

= 6

a n = 205

a + (n -1) d = 205

7 + (n - 1) 6 = 205

(n - 1) 6 = 205 - 7

(n - 1) 6 = 198

n - 1 = 198/6

n - 1 = 33

n = 33 + 1

n = 34

Therefore total number of terms is 34

In this topic 10th CBSE math solution for exercise 5.2 part 5 we are going to see solution of 9th question

(9) If the 3rd and 9th terms of an AP are 4 and -8 respectively,which term of this AP is zero?

Solution:

a₃ = 4

a₉ = -8

a + 2 d = 4 ------- (1)

a + 8 d = -8 ------- (2)

(1) - (2)

a + 2 d = 4

a + 8 d = -8

(-) (-) (+)

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- 6 d = 12

d = -2

Substitute d = -2 in first equation

a + 2 (-2) = 4

a - 4 = 4

a = 4 + 4

a = 8

let us assume nth term is zero

a n = 0

a + (n - 1) d = 0

8 + (n - 1) (-2) = 0

(n - 1) (-2) = -8

n - 1 = 4

n = 5

So fifth term of the sequence is zero.

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(Based on The College Panda's SAT Math : Advanced Guide and Workbook for the New SAT)